Find the length of tangent drawn to a circle with radius 8 cm from a point 17 cm away from the center of the circle.
Let us consider a circle with center O and radius 8 cm.
The diagram is given as:
Consider a point A 17 cm away from the center such that OA = 17 cm
A tangent is drawn at point A on the circle from point B such that OB = radius = 8 cm
To Find: Length of tangent AB = ?
As seen OB ⏊ AB
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled ΔAOB, By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OA)2 = (OB)2 + (AB)2
(17)2 = (8)2 + (AB)2
289 = 64 + (AB)2
(AB)2 = 225
AB = 15 cm
∴ The length of the tangent is 15 cm.
A point P is 25 cm away from the center of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Let us consider a circle with center O.
Consider a point P 25 cm away from the center such that OP = 25 cm
A tangent PQ is drawn at point Q on the circle from point P such that PQ = 24 cm
To Find : Length of radius OQ = ?
Now, OQ ⏊ PQ
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △POQ,
By Pythagoras Theorem,
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OP)2 = (OQ)2 + (PQ)2
(25)2 = (OQ)2 + (24)2
625 = (OQ)2 + 576
(OQ)2 = 49
OQ = 7 cm
Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Given: Two concentric circles (say C1 and C2) with common center as O and radius r1 = 6.5 cm and r2 = 2.5 cm respectively.
To Find: Length of the chord of the larger circle which touches the circle C2. i.e. Length of AB.
As AB is tangent to circle C2 and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
We have,
(OP)2 + (PB)2 = (OB)2
r22 + (PB)2 = r12
(2.5)2 + (PB)2= (6.5)2
6.25 + (PB)2 = 42.25
(PB)2 = 36
PB = 6 cm
Now, AP = PB ,
[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]
So,
AB = AP + PB = PB + PB
= 2PB = 2(6)
= 12 cm
In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.
Let AD = x cm, BE = y cm and CF = z cm
As we know that,
Tangents from an external point to a circle are equal,
In given Figure we have
AD = AF = x [Tangents from point A]
BD = BE = y [Tangents from point B]
CF = CE = z [Tangents from point C]
Now, Given: AB = 12 cm
AD + BD = 12
x + y = 12
y = 12 – x…. [1]
and BC = 8 cm
BE + EC = 8
y + z = 8
12 - x + z = 8 [From 1]
z = x – 4…. [2]
and
AC = 10 cm
AF + CF = 10
x + z = 10 [From 2]
x + x - 4 = 10
2x = 14
x = 7 cm
Putting value of x in [1] and [2]
y = 12 - 7 = 5 cm
z = 7 - 4 = 3 cm
So, we have AD = 7 cm, BE = 5 cm and CF = 3 cm
In the given figure, PA and PB are the tangent segments to a circle with center 0. Show that the points A, O, B and P are concyclic.
Given: PA and PB are tangents to a circle with center O
To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.
Proof:
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic]
Hence Proved.
In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.
Given: Two concentric circles with common center as O
To Prove: AC = CB
Construction: Join OC, OA and OB
Proof :
OC ⏊ AB
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In △OAC and △OCB, we have
OA = OB
[∵ radii of same circle]
OC = OC
[∵ common]
∠OCA = ∠OCB
[∵ Both 90° as OC ⏊ AB]
△OAC ≅ △OCB
[By Right Angle - Hypotenuse - Side]
AC = CB
[Corresponding parts of congruent triangles are congruent]
Hence Proved.
From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of △PCD.
Given : From an external point P, two tangents, PA and PB are drawn to a circle with center O. At a point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. And PA = 14 cm
To Find : Perimeter of △PCD
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
AC = CE …[1] [Tangents from point C]
ED = DB …[2] [Tangents from point D]
Now Perimeter of Triangle PCD
= PC + CD + DP
= PC + CE + ED + DP
= PC + AC + DB + DP [From 1 and 2]
= PA + PB
Now,
PA = PB = 14 cm as tangents drawn from an external point to a circle are equal
So we have
Perimeter = PA + PB = 14 + 14 = 28 cm
A circle is inscribed in a LABC touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
As we know that tangents drawn from an external point to a circle are equal ,
In the Given image we have,
AP = AR = 7 cm ….[1]
[tangents from point A]
CR = QC = 5 cm ….[2]
[tangents from point C]
BQ = PB …[3]
[tangents from point B]
Now,
AB = 10 cm [Given]
AP + PB = 10 cm
7 + PB = 10 [From 1]
PB = 3 cm
BQ = 3 cm …..[4]
[From 3]
BC = BQ + QC = 5 + 3 = 8 cm [ From 2 and 4]
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.
As we know that tangents drawn from an external point to a circle are equal,
In the given image we have,
AP = AS = w (say) [Tangents from point A]
BP = BQ = x (say) [Tangents from point B]
CP = CR = y (say) [Tangents from point C]
DR = DS = z (say) [Tangents from point D]
Now,
Given,
AB = 6 cm
AP + BP = 6
w + x = 6 …[1]
BC = 7 cm
BP + CP = 7
x + y = 7 ….[2]
CD = 4 cm
CR + DR = 4
y + z = 4 ….[3]
Also,
AD = AS + DS = w + z ….[4]
Add [1] and [3] and substracting [2] from the sum we get,
w + x + y + z - (x + y) = 6 + 4 - 7
w + z = 3 cm ; From [4]
AD = 3 cm
In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
As we know that tangents drawn from an external point to a circle are equal,
BR = BP [ Tangents from point B] [1]
QC = CP [ Tangents from point C] [2]
AR = AQ [ Tangents from point A] [3]
As ABC is an isosceles triangle,
AB = BC [Given] [4]
Now substract [3] from [4]
AB - AR = BC - AQ
BR = QC
BP = CP [ From 1 and 2]
∴ P bisects BC
Hence Proved.
In the given figure, O is the center of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB up to one place of decimal.
In given Figure,
OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OAP,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(OP)2 = (OA)2 + (PA)2
Given, PA = 10 cm and OA = radius of outer circle = 6 cm
(OP)2 = (6)2 + (100)2
(OP)2 = 36 + 100 = 136 [1]
Also,
OB ⏊ BP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OBP,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(OP)2 = (OB)2 + (PB)2
Now, OB = radius of inner circle = 4 cm
And from [2]
(OP)2 = 136
136 = (4)2 + (PB)2
(PB)2 = 136 - 16 = 120
PB = 10.9 cm
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ABC = 54 cm2 then find the lengths of sides AB and AC.
Given : △ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm
Also, area(△ABC) = 54 cm2
To Find : AB and AC
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
FB = BD = 6 cm [Tangents from same external point B]
DC = EC = 9 cm [Tangents from same external point C]
AF = EA = x (let) [Tangents from same external point A]
Using the above data, we get
AB = AF + FB = x + 6 cm
AC = AE + EC = x + 9 cm
BC = BD + DC = 6 + 9 = 15 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
Where,
⇒
So for △ABC
a = AB = x + 6
b = AC = x + 9
c = BC = 15 cm
⇒
And
Squaring both sides, we get,
54(54) = 54x(x + 15)
x2 + 15x - 54 = 0
x2 + 18x - 3x - 54 = 0
x(x + 18) - 3(x + 18) = 0
(x - 3)(x + 18) = 0
x = 3 or - 18
but x = - 18 is not possible as length can't be negative.
So
AB = x + 6 = 3 + 6 = 9 cm
AC = x + 9 = 3 + 9 = 12 cm
PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T
To Find : Length of TP
Construction : Join OQ
Now in △OPT and △OQT
OP = OQ [radii of same circle]
PT = PQ
[tangents drawn from an external point to a circle are equal]
OT = OT [Common]
△OPT ≅ △OQT [By Side - Side - Side Criterion]
∠POT = ∠OQT
[Corresponding parts of congruent triangles are congruent]
or ∠POR = ∠OQR
Now in △OPR and △OQR
OP = OQ [radii of same circle]
OR = OR [Common]
∠POR = ∠OQR [Proved Above]
△OPR ≅ △OQT [By Side - Angle - Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Now,
∠ORP + ∠ORQ = 180° [Linear Pair]
∠ORP + ∠ORP = 180°
∠ORP = 90°
⇒ OR ⏊ PQ
⇒ RT ⏊ PQ
As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have
∴ In right - angled △OPR,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(OP)2 = (OR)2 + (PR)2
(3)2 = (OR)2 + (2.4)2
9 = (OR)2 + 5.76
(OR)2= 3.24
OR = 1.8 cm
Now,
In right angled △TPR,
By Pythagoras Theorem
(PT)2 = (PR)2 + (TR)2 …[1]
Also, OP ⏊ OT
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In right angled △OPT, By Pythagoras Theorem
(PT)2 + (OP)2 = (OT)2
(PR)2 + (TR)2 + (OP)2= (TR + OR)2 …[From 1]
(2.4)2 + (TR)2 + (3)2 = (TR + 1.8)2
4.76 + (TR)2 + 9 = (TR)2 + 2(1.8)TR + (1.8)2
13.76 = 3.6TR + 3.24
3.6TR = 10.52
TR = 2.9 cm [Appx]
Using this in [1]
PT2 = (2.4)2 + (2.9)2
PT2 = 4.76 + 8.41
PT2 = 13.17
PT = 3.63 cm [Appx]
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its center.
Given: A circle with center O and AB and CD are two parallel tangents at points P and Q on the circle.
To Prove: PQ passes through O
Construction: Draw a line EF parallel to AB and CD and passing through O
Proof :
∠OPB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
Now, AB || EF
∠OPB + ∠POF = 180°
90° + ∠POF = 180°
∠POF = 90° …[1]
Also,
∠OQD = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
Now, CD || EF
∠OQD + ∠QOF = 180°
90° + ∠QOF = 180°
∠QOF = 90° [2]
Now From [1] and [2]
∠POF + ∠QOF = 90 + 90 = 180°
So, By converse of linear pair POQ is a straight Line
i.e. O lies on PQ
Hence Proved.
In the given figure, a circle with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, LB = 90° and DS = 5 cm then find the radius of the circle.
In quadrilateral POQB
∠OPB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OQB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠PQB = 90° [Given]
By angle sum property of quadrilateral PQOB
∠OPB + ∠OQB + ∠PBQ + ∠POQ = 360°
90° + 90° + 90° + ∠POQ = 360°
∠POQ = 90°
As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle
Also, OP = OQ = r
i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square
∴ POQB is a square
And OP = PB = BQ = OQ = r [1]
Now,
As we know that tangents drawn from an external point to a circle are equal
In given figure, We have
DS = DR = 5 cm
[Tangents from point D and DS = 5 cm is given]
AD = 23 cm [Given]
AR + DR = 23
AR + 5 = 23
AR = 18 cm
Now,
AR = AQ = 18 cm
[Tangents from point A]
AB = 29 cm [Given]
AQ + QB = 29
18 + QB = 29
QB = 11 cm
From [1]
QB = r = 11 cm
Hence Radius of circle is 11 cm.
In the given figure, O is the center of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.
In Given Figure, we have a circle with center O let the radius of circle be r.
Construction : Join OP
Now, In △APB
∠ABP = 30°
∠APB = 90°
[Angle in a semicircle is a right angle]
By angle sum Property of triangle,
∠ABP + ∠APB + ∠PAB = 180
30° + 90° + ∠PAB = 180
∠PAB = 60°
OP = OA = r [radii]
∠PAB = ∠OPA = 60°
[Angles opposite to equal sides are equal]
By angle sum Property of triangle
∠OPA + ∠OAP + ∠AOP = 180°
60° + ∠PAB + ∠AOP = 180
60 + 60 + ∠AOP = 180
∠AOP = 60°
As all angles of △OPA equals to 60°, △OPA is an equilateral triangle
So, we have, OP = OA = PA = r [1]
∠OPT = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OPA + ∠APT = 90
60 + ∠APT = 90
∠APT = 30°
Also,
∠PAB + ∠PAT = 180° [Linear pair]
60° + ∠PAT = 180°
∠PAT = 120°
In △APT
∠APT + ∠PAT + ∠PTA = 180°
30° + 120° + ∠PTA = 180°
∠PTA = 30°
So,
We have
∠APT = ∠PTA = 30°
AT = PA
[Sides opposite to equal angles are equal]
AT = r [From 1] [2]
Now,
AB = OA + OB = r + r = 2r [3]
From [2] and [3]
AB : AT = 2r : r = 2 : 1
Hence Proved !
In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD.
Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.
As we know that tangents drawn from an external point to a circle are equal ,
In the given image we have,
AP = AS = w (say) [Tangents from point A]
BP = BQ = x (say) [Tangents from point B]
CP = CR = y (say) [Tangents from point C]
DR = DS = z (say) [Tangents from point D]
Now,
Given,
AB = 6 cm
AP + BP = 6
w + x = 6 [1]
BC = 9 cm
BP + CP = 9
x + y = 9 [2]
CD = 8 cm
CR + DR = 8
y + z = 8 [3]
Also,
AD = AS + DS = w + z [4]
Add [1] and [3] and substracting [2] from the sum we get,
w + x + y + z - (x + y) = 6 + 8 - 9
w + z = 5 cm
From [4]
AD = 5 cm
In the given figure, PA and PB are two tangents to the circle with center O. If ∠APB = 50° then what is the measure of ∠OAB.
In the given figure, PA and PB are two tangents from common point P
∴ PA = PB
[Tangents drawn from an external point are equal]
∠PBA = ∠PAB
[Angles opposite to equal angles are equal] [1]
By angle sum property of triangle in △APB
∠APB + ∠PBA + ∠PAB = 180°
50° + ∠PAB + ∠PAB = 180° [From 1]
2∠PAB = 130°
∠PAB = 65° [2]
Now,
∠OAP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAB + ∠PAB = 90°
∠OAB + 65° = 90° [From 2]
∠OAB = 25°
In the given figure, O is the center of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.
Given: In the figure, PT and PQ are two tangents and ∠TPQ = 70°
To Find: ∠TRQ
Construction: Join OT and OQ
In quadrilateral OTPQ
∠OTP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OQP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠TPQ = 70° [Common]
By Angle sum of Quadrilaterals,
In quadrilateral OTPQ we have
∠OTP + ∠OQP + ∠TPQ + ∠TOQ = 360°
90° + 90° + 70° + ∠TOQ = 360°
250° + ∠TOQ = 360
∠TQO = 110°
Now,
As we Know the angle subtended by an arc at the center is double the angle subtended by it at any
point on the remaining part of the circle.
∴ we have
∠TOQ = 2∠TRQ
110° = 2 ∠TRQ
∠TRQ = 55°
In the given figure, common tangents AB and CD to the two circles with centers O1 and O2 intersect at E. Prove that AB = CD.
Given: AB and CD are two tangents to two circles which intersects at E .
To Prove: AB = CD
Proof:
As
AE = CE …[1]
[Tangents drawn from an external point to a circle are equal]
And
EB = ED …[2]
[Tangents drawn from an external point to a circle are equal]
Adding [1] and [2]
AE + EB = CE + ED
AB = CD
Hence Proved.
If PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.
Given: PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°
To Find: ∠POQ = ?
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 70° = 90°
∠OPQ = 20°
Also,
OP = OQ [Radii of same circle]
∠OQP = ∠OPQ = 20°
[Angles opposite to equal sides are equal]
In △OPQ By Angle sum property of triangles,
∠OPQ + ∠OQP + ∠POQ = 180°
20° + 20° + ∠POQ = 180°
∠POQ = 140°
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ΔABC = 21 cm2 then find the lengths of sides AB and AC.
Given: △ABC that is drawn to circumscribe a circle with radius r = 2 cm and BD = 4 cm DC = 3cm
Also, area(△ABC) = 21 cm2
To Find: AB and AC
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
FB = BD = 4 cm [Tangents from same external point B]
DC = EC = 3 cm [Tangents from same external point C]
AF = EA = x (let) [Tangents from same external point A]
Using the above data, we get
AB = AF + FB = x + 4 cm
AC = AE + EC = x + 3 cm
BC = BD + DC = 4 + 3 = 7 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
Where,
So, for △ABC
a = AB = x + 4
b = AC = x + 3
c = BC = 7 cm
⇒
And
Squaring both sides,
21(21) = 12x(x + 7)
12x2 + 84x - 441 = 0
4x2 + 28x - 147 = 0
As we know roots of a quadratic equation in the form ax2 + bx + c = 0 are,
So roots of this equation are,
but x = - 10.5 is not possible as length can't be negative.
So
AB = x + 4 = 3.5 + 4 = 7.5 cm
AC = x + 3 = 3.5 + 3 = 6.5 cm
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.
Given : Two concentric circles (say C1 and C2) with common center as O and radius r1 = 5 cm and r2 = 3 cm respectively.
To Find : Length of the chord of the larger circle which touches the circle C2. i.e. Length of AB.
As AB is tangent to circle C2 and,
We know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
We have,
(OP)2 + (PB)2 = (OB)2
r22 + (PB)2 = r12
(3)2 + (PB)2= (5)2
9 + (PB)2 = 25
(PB)2 = 16
PB = 4 cm
Now, AP = PB,
[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]
So,
AB = AP + PB = PB + PB
= 2PB = 2(4) = 8 cm
Prove that the perpendicular at the point of contact of the tangent to a circle passes through the center.
Let us consider a circle with center O and XY be a tangent
To prove : Perpendicular at the point of contact of the tangent to a circle passes through the center i.e. the radius OP ⏊ XY
Proof :
Take a point Q on XY other than P and join OQ .
The point Q must lie outside the circle. (because if Q lies inside the circle, XY
will become a secant and not a tangent to the circle).
∴ OQ is longer than the radius OP of the circle. That is,
OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the
shortest of all the distances of the point O to the points of XY.
So OP is perpendicular to XY.
[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]
In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with center 0. If ∠PRQ = 120°, then prove that OR = PR + RQ.
Given : In the figure ,
Two tangents RQ and RP are drawn from an external point R to the circle with center O and ∠PRQ = 120°
To Prove: OR = PR + RQ
Construction: Join OP and OQ
Proof :
In △△OPR and △OQR
OP = OQ [radii of same circle]
OR = OR [Common]
PR = PQ …[1]
[Tangents drawn from an external point are equal]
△OPR ≅ △OQR
[By Side - Side - Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Also,
∠PRQ = 120°
∠ORP + ∠ORQ = 120°
∠ORP + ∠ORP = 120°
2∠ORP = 120°
∠ORP = 60°
Also, OP ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In right angled triangle OPR,
∴ OR = 2PR
OR = PR + PR
OR = PR + RQ [From 1]
Hence Proved.
In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.
Let AD = x cm, BE = y cm and CF = z cm
As we know that,
Tangents from an external point to a circle are equal,
In given Figure we have
AD = AF = x
[Tangents from point A]
BD = BE = y
[Tangents from point B]6CF = CE = z [Tangents from point C]
Now, Given: AB = 14 cm
AD + BD = 14
x + y = 14
y = 14 - x … [1]
and BC = 8 cm
BE + EC = 8
y + z = 8
14 - x + z = 8 … [From 1]
z = x - 6 [2]
and
CA = 12 cm
AF + CF = 12
x + z = 12 [From 2]
x + x - 6 = 12
2x = 18
x = 9 cm
Putting value of x in [1] and [2]
y = 14 - 9 = 5 cm
z = 9 - 6 = 3 cm
So, we have AD = 9 cm, BE = 5 cm and CF = 3 cm
In the given figure, 0 is the center of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.
Given : PA and PB are tangents to a circle with center O
To show : AOBP is a cyclic quadrilateral.
Proof :
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]
Hence Proved.
In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.
Let us consider circles C1 and C2 with common center as O. Let AB be a tangent to circle C1 at point P and chord in circle C2. Join OB
In △OPB
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2]
(OB)2 = (OP)2 + (PB)2
Now, 2PB = AB
[As we have proved above that OP ⏊ AB and Perpendicular drawn from center to a chord bisects the chord]
2PB = 8 cm
PB = 4 cm
(OB)2 = (5)2 + (4)2
[As OP = 5 cm, radius of inner circle]
(OB)2 = 41
⇒ OB = √41 cm
In the given figure, PQ is a chord of a circle with center 0 and PT is a tangent. If ∠QPT = 60°, find ∠P
Given : , PQ is a chord of a circle with center 0 and PT is a tangent and ∠QPT = 60°.
To Find : ∠PRQ
∠OPT = 90°
∠OPQ + ∠QPT = 90°
∠OPQ + 60° = 90°
∠OPQ = 30° … [1]
Also.
OP = OQ [radii of same circle]
∠OQP = ∠OPQ [Angles opposite to equal sides are equal]
From [1], ∠OQP = ∠OQP = 30°
In △OPQ , By angle sum property
∠OQP + ∠OPQ + ∠POQ = 180°
30° + 30° + ∠POQ = 180°
∠POQ = 120°
As we know, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
So, we have
2∠PRQ = reflex ∠POQ
2∠PRQ = 360° - ∠POQ
2∠PRQ = 360° - 120° = 240°
∠PRQ = 120°
In the given figure, PA and PB are two tangents to the circle with center O. If ∠APB = 60° then find the measure of ∠OAB.
In the given figure, PA and PB are two tangents from common point P
∴ PA = PB
[∵ Tangents drawn from an external point are equal]
∠PBA = ∠PAB
[∵ Angles opposite to equal angles are equal] …[1]
By angle sum property of triangle in △APB
∠APB + ∠PBA + ∠PAB = 180°
60° + ∠PAB + ∠PAB = 180° [From 1]
2∠PAB = 120°
∠PAB = 60° …[2]
Now,
∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAB + ∠PAB = 90°
∠OAB + 60° = 90° [From 2]
∠OAB = 30°
The number of tangents that can be drawn from an external point to a circle is
A. 1
B. 2
C. 3
D. 4
The maximum number of tangents that can be drawn from an external point to a circle is two and they are equal in length.
In the given figure, RQ is a tangent to the circle with center O. If SQ = 6 cm and QR = 4 cm, then OR is equal to
A. 2.5 cm
B. 3 cm
C. 5 cm
D. 8 cm
As SQ is diameter and OQ is radius in the given circle,
∴ 2OQ = SQ [As 2 × radius) = diameter]
2OQ = 6 cm
OQ = 3 cm
Now, QR is tangent
∴ OQ ⏊ QR
In right - angled △OQR,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ]
(QR)2 + (OQ)2 = (OR)2
(4)2 + (3)2 = (OR)2
16 + 9 = (OR)2
(OR)2 = 25
OR = 5 cm
In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If O is the center of the circle, then length OP = ?
A. 30 cm
B. 28 cm
C. 25 cm
D. 18 cm
We have given, PT is a tangent drawn at point T on the circle.
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2]
(OP)2 = (OT)2 + (PT)2
(OP)2 = (7)2 + (24)2
(OP)2 = 49 + 576
(OP)2 = 625
⇒ OP = 25 cm
Which of the following pairs of lines in a circle cannot be parallel?
A. two chords
B. a chord and a tangent
C. two tangents
D. two diameters
As all diameters of a circle passes through center O it is not possible to have two parallel diameters in a circle.
The chord of a circle of radius 10 cm subtends a right angle at its center. The length of the chord (in cm) is
A.
B.
C.
D.
Let us consider a circle with center O and AB be any chord that subtends 90° angle at its center.
Now, In △OAB
OA = OB = 10 cm
And as ∠AOB = 90° ,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2]
(OA)2 + (OB)2 = (AB)2
(10)2+ (10)2 = (AB)2
100 + 100 = (AB)2
⇒ AB = √200 = 10√2
So, Correct option is C .
In the given figure, PT is a tangent to the circle with center O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is
A. 8 cm
B. 10 cm
C. 12 cm
D. 16 cm
We have given, PT is a tangent drawn at point T on the circle.
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2]
(OP)2 = (OT)2 + (PT)2
(10)2 = (6)2 + (PT)2
(PT)2 = 100 - 36
(PT)2 = 64
⇒ PT = 8 cm
In the given figure, point P is 26 cm away from the center 0 of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then, the radius of the circle is
A. 10 cm
B. 12 cm
C. 13 cm
D. 15 cm
We have given, PT is a tangent drawn at point T on the circle and OP = 26 cm and PT = 24 cm
Join OT
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2]
(OP)2 = (OT)2 + (PT)2
(26)2 = (OT)2 + (24)2
(OT)2 = 676 - 576
(OT)2 = 100
OT = 10 cm
Hence, radius of circle is 10 cm.
PQ is a tangent to a circle with center O at the point P. If ΔOPQ is an isosceles triangle, then ∠OQP is equal to
A. 30°
B. 45°
C. 60°
D. 90°
Let us consider a circle with center O and PQ is a tangent
on the circle, Joined OP and OQ
But OPQ is an isosceles triangle, ∴ OP = PQ
∠OQP = ∠POQ
[Angles opposite to equal sides are equal]
In △OQP
∠OQP + ∠OPQ + ∠POQ = 180°
[Angle sum property of triangle]
∠OQP + 90° + ∠OPQ = 180°
2 ∠OPQ = 90°
∠OPQ = 45°
In the given figure, AB and AC are tangents to the circle with center O such that ∠BAC = 40°. Then, ∠BOC is equal to
A. 80°
B. 100°
C. 120°
D. 140°
As AB and AC are tangents to given circle,
We have,
OB ⏊ AB and OC ⏊ AC
[∵ Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OBA = ∠OCA = 90°
In quadrilateral AOBC, By angle sum property of quadrilateral, we have,
∠OBA + ∠OCA + ∠BOC + ∠BAC = 360°
90° + 90° + ∠BOC + 40° = 360°
∠BOC = 140°
If a chord AB subtends an angle of 60° at the center of a circle, then the angle between the tangents to the circle drawn from A and B is
A. 30°
B. 60°
C. 90°
D. 120°
Let us consider a circle with center O and AB be a chord such that ∠AOB = 60°
AP and BP are two intersecting tangents at point P at point A and B respectively on the circle.
To find : Angle between tangents, i.e. ∠APB
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP, By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + 60° = 360°
∠APB = 120°
In the given figure, O is the center of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of chord AB is
A. 8 cm
B. 14 cm
C. 16 cm
D.
Given: Two concentric circles (say C1 and C2) with common center as O and radius r1 = 6 cm(inner circle) and r2 = 10 cm (outer circle) respectively.
To Find : Length of the chord AB.
As AB is tangent to circle C1 and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
We have,
(OP)2 + (PA)2 = (OA)2
r12 + (PA)2 = r22
(6)2 + (PA)2= (10)2
36 + (PA)2 = 100
(PA)2 = 64
PA = 8 cm
Now, PA = PB ,
[ as perpendicular from center to chord bisects the chord and OP ⏊ AB]
So,
AB = PA + PB = PA + PA = 2PA = 2(8) = 16 cm
In the given figure, AB and AC are tangents to a circle with center 0 and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is
A. 9
B. 15
C. 353
D. 25
As AB is tangent to the circle at point B
OB ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
In right angled triangle AOB,
By Pythagoras Theorem,
[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ]
(OA)2 = (OB)2 + (AB)2
(17)2 = (8)2 + (AB)2
[As OA = 17 cm is given and OB is radius]
289 = 64 + (AB)2
(AB)2 = 225
AB = 15 cm
Now, AB = AC [Tangents drawn from an external point are equal]
∴ AC = 15 cm
In the given figure, 0 is the center of a circle, AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A then ∠BAT = ?
A. 40°
B. 50°
C. 60°
D. 65°
In △ABC
∠ABC = 90°
[Angle in a semicircle is a right angle]
∠ACB = 50° [Given]
By angle sum Property of triangle,
∠ACB + ∠ABC + ∠CAB = 180°
90° + 50° + ∠CAB = 180°
∠CAB = 40°
Now,
∠CAT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠CAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 50°
In the given figure, O is the center of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then ∠TPQ is equal to
A. 35°
B. 45°
C. 55°
D. 70°
In △OPQ
∠POQ = 70° [Given]
OP = OQ [radii of same circle]
∠OQP = ∠OPQ [Angles opposite to equal sides are equal]
By angle sum Property of triangle,
∠POQ + ∠OQP + ∠OPQ = 180°
70° + ∠OPQ + ∠OPQ = 180°
2 ∠OPQ = 110°
∠OPQ = 55°
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠TPQ = 90°
55° + ∠TPQ = 90°
∠TPQ = 35°
In the given figure, AT is a tangent to the circle with center O such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?
A. 4 cm
B. 2 cm
C. 2√3 cm
D. 4√3 cm
Given: AT is a tangent to the circle with center O such that OT = 4 cm and ∠OTA = 30°.
To find: The value of AT.
Solution:
In △ OAT ,
OA ⏊ AT [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴OAT is a right - angled triangle at A and
AT = 2√3 cm
If PA and PB are two tangents to a circle with center O such that ∠AOB = 110° then ∠APB is equal to
A. 55°
B. 60°
C. 70°
D. 90°
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + 110° + ∠APB = 360°
∠APB = 70°
In the given figure, the length of BC is
A. 7 cm
B. 10 cm
C. 14 cm
D. 15 cm
As we know,
Tangents drawn from an external point are equal, We have
AF = AE = 4 cm
[Tangents from common point A]
BF = BD = 3 cm
[Tangents from common point B]
CE = CD = x (say)
[Tangents from common point C]
Now,
AC = AE + CE
11 = 4 + x
x = 7 cm [1]
and, BC = BD + BC
BC = 3 + x = 3 + 7 = 10 cm
In the given figure, if ∠AOD = 135° then ∠BOC is equal to
A. 25°
B. 45°
C. 52.5°
D. 62.5°
In the given figure, 0 is the center of a circle and PT is the tangent to the circle. If PQ is a chord such that ∠QPT = 50° then ∠POQ = ?
A. 100°
B. 90°
C. 80°
D. 75°
In the given figure PT is a tangent to circle ∴ we have
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 50° = 90°
∠OPQ = 40°
Now, In △POQ
OP = OQ
∠PQO = ∠QPO = 40°
[Angles opposite to equal sides are equal]
Now,
∠ PQO + ∠QPO + ∠ POQ = 180°
[By angle sum property of triangle]
40° + 40° + ∠POQ = 180°
∠POQ = 100°
In the given figure, PA and PB are two tangents to the circle with center O. If ∠APB = 60° then ∠OAB is
A. 15°
B. 30°
C. 60°
D. 90°
In the given figure, PA and PB are two tangents from common point P
∴ PA = PB
[Tangents drawn from an external point are equal]
∠PBA = ∠PAB…[1]
[Angles opposite to equal angles are equal]
By angle sum property of triangle in △APB
∠APB + ∠PBA + ∠PAB = 180°
60° + ∠PAB + ∠PAB = 180° [From 1]
2∠PAB = 120°
∠PAB = 60°…[2]
Now,
∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAB + ∠PAB = 90°
∠OAB + 60° = 90° [From 2]
∠OAB = 30°
If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm then the length of each tangent is
A. 3 cm
B.
C.
D. 6 cm
Let us consider a circle with center O and AP and BP are two tangents such that angle of inclination i.e. ∠APB = 60°
Joined OA, OB and OP.
To Find : Length of tangents
Now,
PA = PB [Tangents drawn from an external point are equal] [1]
In △AOP and △BOP
PA = PB [By 1]
OP = OP [Common]
OA = OB [radii of same circle]
△AOP ≅△BOP
[By Side - Side - Side Criterion]
∠OPA = ∠OPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 60° [Given]
∠OPA + ∠OPB = 60°
∠OPA + ∠OPA = 60°
2 ∠OPA = 60°
∠OPA = 30°
In △AOP
OA ⏊ PA
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact
∴ AOP is a right - angled triangle.
So, we have
⟹PA = 3√3 cm
From [1]
PA = PB = 4 cm
i.e. length of each tangent is 3√3 cm
In the given figure, PQ and PR are tangents to a circle with center A. If ∠QPA = 27° then ∠QAR equals
A. 63°
B. 117°
C. 126°
D. 153°
In Given Figure,
PQ = PR…[1]
[Tangents drawn from an external point are equal]
In △AOP and △BOP
PQ = PR [By 1]
AP = AP [Common]
AQ = AR [radii of same circle]
△AQP ≅△ARP [By Side - Side - Side Criterion]
∠QPA = ∠RPA
[Corresponding parts of congruent triangles are congruent]
Now,
∠QPA + ∠RPA = ∠QPR
∠QPA + ∠QPA = ∠QPR
2 ∠QPA = ∠QPR
∠QPR = 2(27) = 54°
As PQ and PQ are tangents to given circle,
We have,
AQ ⏊ PQ and AR ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠AQP = ∠ARP = 90°
In quadrilateral AQRP, By angle sum property of quadrilateral, we have
∠AQP + ∠ARP + ∠QAR + ∠QPR = 360°
90° + 90° + ∠QAR + 54° = 360°
∠QAR = 126°
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with center C and radius 4 cm. If PA ⏊ PB, then the length of each tangent is
A. 3 cm
B. 4 cm
C. 5 cm
D. 6 cm
Join AC, BC and CP
To Find: Length of tangents
Now,
PA = PB…[1]
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
In △ACP and △BCP
PA = PB [By 1]
CP = CP [Common]
CA = CB [radii of same circle]
△ACP ≅△BCP [By Side - Side - Side Criterion]
∠CPA = ∠CPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 90°
[Given that PA ⏊ PB]
∠CPA + ∠CPB = 90°
∠CPA + ∠CPA = 90°
2 ∠CPA = 90°
∠CPA = 45°
In △ACP
CA ⏊ PA [Tangents drawn at a point on circle is perpendicular to the radius through point of contact
∴ ACP is a right - angled triangle.
So, we have
⟹PA = 4 cm
From [1]
PA = PB = 4 cm
i.e. length of each tangent is 4 cm
If PA and PB are two tangents to a circle with center O such that ∠APB = 80°. Then, ∠AOP = ?
A. 40°
B. 50°
C. 60°
D. 70°
In Given Figure,
PA = PB…[1]
[Tangents drawn from an external point are equal]
In △AOP and △BOP
PA = PB [By 1]
OP = OP [Common]
OA = OB
[radii of same circle]
△AOP ≅△BOP
[By Side - Side - Side Criterion]
∠OPA = ∠OPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 80° [Given]
∠OPA + ∠OPB = 80°
∠OPA + ∠OPA = 80°
2 ∠OPA = 80°
∠OPA = 40°
In △AOP,
∠OAP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
And
∠OAP + ∠OPA + ∠AOP = 180°
90° + 40° + ∠AOP = 180°
∠AOP = 50°
In the given figure, O is the center of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58° then the measure of ∠PQB is
A. 32°
B. 58°
C. 122°
D. 132°
In the given Figure, Join OP
Now, OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴∠OPA = 90°
∠OPQ + ∠APQ = 90°
∠OPQ + 58° = 90°
[Given, ∠APQ = 58°]
∠OPQ = 32°
In △OPQ
OP = OQ
[Radii of same circle]
∠OQP = ∠OPQ
[Angles opposite to equal sides are equal]
∠PQB = 32°
[As ∠OQP = ∠PQB ]
In the given figure, O is the center of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30° then ∠CPB + ∠ACP is equal to
A. 60°
B. 90°
C. 120°
D. 150°
In given Figure, Join OP
In △OPC,
OP = OC [Radii of same circle]
∠OCP = ∠OPC
[Angles opposite to equal sides are equal]
∠ACP = ∠OPC
[As ∠OCP = ∠ACP] …[1]
Now,
∠OPB = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPC + ∠CPB = 90°
∠ACP + ∠CPB = 90° [By 1]
So,
∠CPB + ∠ACP = 90°
In the given figure, PQ is a tangent to a circle with center O. A is the point of contact. If ∠PAB = 67°, then the measure of ∠AQB is
A. 73°
B. 64
C. 53°
D. 44°
In the given Figure, Join OA
Now,
OA ⏊ PQ
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAP = ∠OAQ = 90° [1]
∠OAB + ∠PAB = 90°
∠OAB + 67° = 90°
∠OAB = 23°
Now,
∠BAC = 90°
[Angle in a semicircle is a right angle]
∠OAB + ∠OAC = 90°
23° + ∠OAC = 90°
∠OAC = 67°
∠OAQ = 90° [From 1]
∠OAC + ∠CAQ = 90°
67° + ∠CAQ = 90°
∠CAQ = 23° [2]
Now,
OA = OC
[radii of same circle]
∠OCA = ∠OAC
[Angles opposite to equal sides are equal]
∠OCA = 67°
∠OCA + ∠ACQ = 180° [Linear Pair]
67° + ∠ACQ = 180°
∠ACQ = 113° [3]
Now, In △ACQ By Angle Sum Property of triangle
∠ACQ + ∠CAQ + ∠AQC = 180°
113° + 23° + ∠AQC = 180° [By 2 and 3]
∠AQC = 44°
In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is
A. 45°
B. 60°
C. 90°
D. 120°
O is the center of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to P the circle. Then, the area of quad. PQOR is
A. 60 cm2
B. 32.5 cm2
C. 65 cm2
D. 30 cm2
In Given Figure,
PQ = PR…[1]
[Tangents drawn from an external point are equal]
In △QOP and △ROP
PQ = PR [By 1]
OP = OP [Common]
OQ = OR [radii of same circle]
△QOP ≅△ROP
[By Side - Side - Side Criterion]
area(Δ QOP) = area(Δ ROP)
[Congruent triangles have equal areas]
area(PQOR) = area(Δ QOP) + area(Δ ROP)
area(PQOR) = area(Δ QOP) + area(Δ QOP) = 2[area(Δ QOP)]
Now,
OQ ⏊PQ
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, QOP is a right - angled triangle at Q with OQ as base and PQ as height.
In △QOP,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(OQ)2 + (PQ)2 = (OP)2
(5)2 + (PQ)2 = (13)2
25 + (PQ)2 = 169
(PQ)2 = 144
PQ = 12 cm
Area(ΔQOP) = 1/2 × Base × Height
= 1/2 × OQ × PQ
= 1/2 × 5 × 12
= 30 cm2
So,
Area(PQOR) = 2(30) = 60 cm2
In the given figure, PQR is a tangent to the circle at Q, whose center is O and AB is a chord parallel to PR such that ∠BQR = 70°. Then, ∠AQB =?
A. 20°
B. 35°
C. 40°
D. 45°
In given figure, as PR is a tangent
OQ ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
⟹LQ ⏊ PR
⟹LQ ⏊ AB
[As, AB || PR]
AL = LB
[Perpendicular from center to the chord bisects the chord]
Now,
∠LQR = 90°
∠LQB + ∠BQR = 90°
∠LQB + 70° = 90°
∠LQB = 20°…[1]
In △AQL and △BQL
∠ALQ = ∠BLQ [Both 90° as LQ ⏊ AB]
AL = LB [Proved above]
QL = QL [Common]
△AQL ≅ △BQL
[Side - Angle - Side Criterion]
∠LQA = ∠LQB
[Corresponding parts of congruent triangles are congruent]
∠AQB = ∠LQA + ∠LQB = ∠LQB + ∠LQB
= 2∠LQB = 2(20) = 40° [By 1]
The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the center of the circle is
A. 8 cm
B. √104 cm
C. 12 cm
D. √125 cm
Let us consider a circle with center O and TP be a tangent at point A on the circle, Joined OT and OP
Given Length of tangent, TP = 10 cm, and OT = 5 cm [radius]
To Find : Distance of center O from P i.e. OP
Now,
OP ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So OPT is a right - angled triangle,
By Pythagoras Theorem in ΔOPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OT)2 + (TP)2 = (OP)2
(OP)2 = (5)2 + (10)2
(OP)2 = 25 + 100 = 125
OP = √125 cm
In the given figure, 0 is the center of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30° then ∠PTA = ?
A. 60°
B. 30°
C. 15°
D. 45°
In △BOP
OB = OP [radii of same circle]
∠OPB = ∠PBO
[Angles opposite to equal sides are equal]
As, ∠PBO = 30°
∠OPB = 30°
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠BPT = ∠OPB + ∠OPT = 30° + 90° = 120°
Now, In ΔBPT
∠BPT + ∠PBO + ∠PTB = 180°
120° + 30° + ∠PTB = 180°
∠PTB = 30°
∠PTA = ∠PTB = 30°
In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of ΔEDF is
A. 9 cm
B. 12 cm
C. 13.5 cm
D. 18 cm
Given : In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively and EK = 9 cm
To Find : Perimeter of △EDF
As we know that, Tangents drawn from an external point to a circle are equal.
So, we have
KD = DH …[1]
[Tangents from point D]
HF = FM …[2]
[Tangents from point F]
Now Perimeter of Triangle PCD
= ED + DF + EF
= ED + DH + HF + EF
= ED + KD + FM + EF [From 1 and 2]
= EK + EM
Now,
EK = EM = 9 cm as tangents drawn from an external point to a circle are equal
So, we have
Perimeter = EK + EM = 9 + 9 = 18 cm
To draw a pair of tangents to a circle, which is inclined to each other at an angle of 45°, we have to draw tangents at the end points of those two radii, the angle between which is
A. 105°
B. 135°
C. 140°
D. 145°
Let us consider a circle with center O and PA and PB are two tangents from point P, given that angle of inclination i.e. ∠APB = 45°
As PA and PB are tangents to given circle,
We have,
OA ⏊ PA and OB ⏊ PB [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AQRP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠ABP + ∠AOB = 360°
90° + 90° + 45° + ∠AOB = 360°
∠AOB = 135°
In the given figure, O is the center of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, ∠QSR = ?
A. 40°
B. 50°
C. 60°
D. 70°
As PL and PM are tangents to given circle,
We have,
OR ⏊ PM and OQ ⏊ PL
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠ORM = ∠OQL = 90°
∠ORM = ∠ORS + ∠SRM
90° = ∠ORS + 60°
∠ORS = 30°
And ∠OQL = ∠OQS + ∠SQL
90° = ∠OQS + 50°
∠OQS = 40°
Now, In △SOR
OS = OQ [radii of same circle]
∠ORS = ∠OSR
[Angles opposite to equal sides are equal]
∠OSR = 30°
[as ∠ORS = 30°]
Now, In △SOR
OS = SQ [radii of same circle]
∠OQS = ∠OSQ
[Angles opposite to equal sides are equal]
∠OSQ = 40° [as ∠OQS = 40°]
As,
∠QSR = ∠OSR + ∠OSQ
∠QSR = 30° + 40° = 70°
In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of ΔPQR = 189 cm2 then the length of side PQ is
A. 17.5 cm
B. 20 cm
C. 22.5 cm
D. 25cm
Given : △PQR that is drawn to circumscribe a circle with radius r = 6 cm and QT = 12 cm QR = 9cm
Also, area(△PQR) = 189 cm2
Let tangents PR and PQ touch the circle at X and Y respectively.
To Find : PQ and QR
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
QT = QY = 12 cm
[Tangents from same external point B]
TR = RX = 9 cm
[Tangents from same external point C]
PX = PY = x (let)
[Tangents from same external point A]
Using the above data we get
PQ = PY + QT = x + 12 cm
PR = PC + RX = x + 9 cm
QR = QT + TR = 12 + 9 = 21 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
Where,
So for △PQR
a = PQ = x + 12
b = PR = x + 9
c = QR = 21 cm
And
Squaring both side
189(189) = 108(x + 21)
7(189) = 4(x + 21)
4x2 + 84x - 1323 = 0
As we know roots of a quadratic equation in the form ax2 + bx + c = 0 are,
So, roots of this equation are,
but x = - 31.5 is not possible as length can't be negative.
So
PQ = x + 12 = 10.5 + 12 = 22.5 cm
In the given figure, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is
A. 1.9 cm
B. 3.8 cm
C. 5.7 cm
D. 7.6 cm
Let the bigger circle be C1 and Smaller be C2,
Now,
PQ and PT are two tangents to circle C1,
∴ PT = QP
[Tangents drawn from an external point are equal]
QP = 3.8 cm
[ As PT = 3.8 cm is given]
Also,
PR and PT are two tangents to circle C2,
∴ PT = PR
[Tangents drawn from an external point are equal]
PR = 3.8 cm
[ As PT = 3.8 cm is given]
QR = QP + PR = 3.8 + 3.8 = 7.6 cm
In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm. Then, the length AB = ?
A. 9 cm
C. 12 cm
B. 10 cm
D. 8 cm
As we know Tangents drawn from an external point are equal]
In the given Figure, we have
AP = AQ = 5 cm
[Tangents from point A] [AP = 5 cm is given]
BQ = BR = x(say)
[Tangents from point B]
CR = CS = 3 cm [∵ CS = 3 cm is given]
[Tangents from point C]
Given,
BC = 7 cm
CR + BR = 7
3 + x = 7 cm
x = 4 cm
Now,
AB = AQ + BQ = 5 + x = 5 + 4 = 9 cm
In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm then perimeter of quad. ABCD is
A. 18 cm
C. 36 cm
B. 27 cm
D. 32 cm
As we know Tangents drawn from an external point are equal]
In the given Figure, we have
AP = AS = 6 cm [AP = 6 cm is given]
[∵ Tangents from point A]
BP = BQ = 5 cm [BP = 5 cm is given]
[∵ Tangents from point B]
CR = CQ = 3 cm [CQ = 3 cm is given]
[∵ Tangents from point C]
DR = DS = 4 cm ][DR = 4 cm is given]
[∵ Tangents from point D]
Now,
Perimeter of ABCD = AB + BC + CA + DA
= AP + BP + BQ + CQ + CR + DR + DS + AS
= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6 = 36 cm
In the given figure, O is the center of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100° then ∠BAT is equal to
A. 40°
C. 90°
B. 50°
D. 100°
In △AOB
OA = OB [radii of same circle]
∠OBA = ∠OAB [Angles opposite to equal sides are equal]
Also, By Triangle sum Property
∠AOB + ∠OBA + ∠OAB = 180°
100 + ∠OAB + ∠OAB = 180°
2 ∠OAB = 90°
∠OAB = 40°
As AT is tangent to given circle,
We have,
OA ⏊ AT
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAT = 90°
∠OAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 50°
In a right triangle ABC, right - angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is
A. 1 cm
B. 2 cm
C. 3 cm
D. 4 cm
Let AB, BC and AC touch the circle at points P, Q and R respectively.
As ABC is a right triangle,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(AC)2 = (BC)2 + (AB)2
(AC)2 = (12)2 + (5)2
(AC)2 = 144 + 25 = 169
AC = 13 cm
Let O be the center of circle, Join OP, OQ and PR
Let the radius of circle be r,
We have
r = OP = OQ = OR
[radii of same circle] [1]
Now,
ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)
As we know,
Area of triangle is 1/2 × Base × Height (Altitude)
Now,
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OP is the altitude in △AOB
OQ ⏊ BC
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OQ is the altitude in △BOC
OR ⏊ AC
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OR is the altitude in △AOC
So, we have
1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)
12(5) = 5(r) + 12(r) + 13(r) [Using 1]
60 = 30r
r = 2 cm
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. lithe radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD⏊ CD then the length of CD is
A. 11 cm
B. 15 cm
C. 20 cm
D. 21 cm
In quadrilateral ORDS
∠ORD = 90°
[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OSD = 90°
[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠SDR = 90° [AD ⏊ CD]
By angle sum property of quadrilateral PQOB
∠ORD + ∠OSD + ∠SDR + ∠SOR = 360°
90° + 90° + 90° + ∠SOR = 360°
∠SOR = 90°
As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle
Also, OS = OR = r
i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square
∴ POQB is a square
And OS = OR = DR = DS = r = 10 cm [1]
Now,
As we know that tangents drawn from an external point to a circle are equal
In given figure, We have
CQ = CR …[2]
[∵ tangents from point C]
PB = BQ = 27 cm
[∵Tangents from point B and PB = 27 cm is given]
BC = 38 cm [Given]
BQ + CQ = 38
27 + CQ = 38
CQ = 11 cm
From [2]
CQ = CR = 11 cm
Now,
CD = CR + DR
CD = 11 + 10 = 21 cm [from 1, DR = 10 cm]
In the given figure, LABC is right - angled at B such that BC = 6 cm and AB = 8 cm. A circle with center O has been inscribed inside the triangle. OP ⊥AB, OQ ⊥BC and OR ⊥AC. If OP = OQ = OR = x cm then x = ?
A. 2 cm
B. 2.5 cm
C. 3 cm
D. 3.5 cm
As ABC is a right triangle,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(AC)2 = (BC)2 + (AB)2
(AC)2 = (6)2 + (8)2
(AC)2 = 36 + 64 = 100
AC = 10 cm
Now,
ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)
As we know,
Area of triangle is 1/2 × Base × Height(Altitude)
Now,
OP ⏊ AB [Given]
∴ OP is the altitude in △AOB
OQ ⏊ BC [Given]
∴ OQ is the altitude in △BOC
OR ⏊ AC [Given]
∴ OR is the altitude in △AOC
So, we have
1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)
6(8) = 8(x) + 6(x) + 10(x)
[∵ OP = OQ = OR = x, Given]
48 = 24x
x = 2 cm
Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm then the length of AD is
A. 3 cm
B. 4 cm
C. 6 cm
D. 7 cm
Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.
As we know that tangents drawn from an external point to a circle are equal,
So, we have,
AP = AS = w (say)
[∵ Tangents from point A]
BP = BQ = x (say)
[∵Tangents from point B]
CP = CR = y (say)
[∵Tangents from point C]
DR = DS = z (say)
[∵Tangents from point D]
Now,
Given,
AB = 6 cm
AP + BP = 6
w + x = 6 …[1]
BC = 7 cm
BP + CP = 7
x + y = 7 …[2]
CD = 4 cm
CR + DR = 4
y + z = 4 …[3]
Also,
AD = AS + DS = w + z …[4]
Add [1] and [3] and substracting [2] from the sum we get,
w + x + y + z - (x + y) = 6 + 4 - 7
w + z = 3 cm
From [4]
AD = 3 cm
In the given figure, DE and DF are tangents from an external point D to a circle with center A. If DE = 5 cm and DE ⊥ DF then the radius of the circle is
A. 3 cm
B. 4 cm
C. 5 cm
D. 6 cm
.
In the given figure, three circles with centers A, B, C respectively touch each other externally.
If AB = 5 cm, BC = 7 cm and CA = 6 cm then the radius of the circle with center A is
A. 1.5 cm
B. 2 cm
C. 2.5 cm
D. 3 cm
In the given Figure
AR = AP = x(let) [Radii of same circle]
BP = BQ = y(let) [Radii of same circle]
CR = CQ = z(let) [Radii of same circle]
Now,
AB = 5 cm [Given]
AP + BP = 5
x + y = 5
y = 5 - x …[1]
BC = 7 cm [Given]
BQ + CQ = 7
y + z = 7
5 - x + z = 7 [using 1]
z = 2 + x …[2]
and
AC = 6 cm [Given]
x + z = 6
x + 2 + x = 6 [Using 2]
2x = 4
x = 2 cm
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is
A. 15 cm
B. 10 cm
C. 9 cm
D. 7.5 cm
Let tangent BC touch the circle at point R
As we know tangents drawn from an external point to a circle are equal.
We have
AP = AQ
[tangents from point A]
BP = BR …[1]
[tangents from point B]
CQ = CR …[2]
[tangents from point C]
Now,
AP = AQ
⇒ AB + BP = AC + CQ
⇒ 5 + BR = 6 + CR [From 1 and 2]
⇒CR = BR - 1 …[3]
Now,
BC = 4 cm
BR + CR = 4
BR + BR - 1 = 4 [Using 3]
2BR = 5 cm
BR = 2.5 cm
BP = BR = 2.5 cm [Using 2]
AP = AB + BP = 5 + 2.5 = 7.5 cm
In the given figure, O is the center of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to
A. 5 cm
B. 3√5 cm
C. 4√10 cm
D. 5√10 cm
In given Figure,
OA ⏊ AP [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OAP,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]
(OP)2 = (OA)2 + (PA)2
Given, PA = 12 cm and OA = radius of outer circle = 5 cm
(OP)2 = (5)2 + (12)2
(OP)2 = 25 + 144 = 136
OP = 13 cm …[1]
Also,
OB ⏊ BP [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OBP,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OP)2 = (OB)2 + (PB)2
Now, OB = radius of inner circle = 3 cm
And, from [2] (OP) = 13 cm
(13)2 = (3)2 + (PB)2
(PB)2 = 169 - 9 = 160
PB = 4√10 cm
Which of the following statements is not true?
A. If a point P lies inside a circle, no tangent can be drawn to the circle, passing through P.
B. If a point P lies on the circle, then one and only one tangent can be drawn to the circle at P.
C. If a point P lies outside the circle, then only two tangents can be drawn to the circle from P.
D. A circle can have more than two parallel tangents, parallel to a given line.
A circle cannot have more than two tangents parallel, because tangents to be parallel they should be at diametrically ends and a diameter has two ends only.
Which of the following statements is not true?
A. A tangent to a circle intersects the circle exactly at one point.
B. The point common to the circle and its tangent is called the point of contact.
C. The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
D. A straight line can meet a circle at one point only.
A straight line can meet a circle at two points in case if it is a chord or diameter or a line intersecting the circle at two points.
Which of the following statements is not true?
A. A line which intersects a circle in two points, is called a secant of the circle.
B. A line intersecting a circle at one point only, is called a tangent to the circle.
C. The point at which a line touches the circle, is called the point of contact.
D. A tangent to the circle can be drawn from a point inside the circle.
If a tangent is drawn from a point inside a circle, it will intersect the circle at two points, so no tangent can be drawn from a point inside the circle.
Assertion - and - Reason Type
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A)
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
Let us consider a circle with center O and radius 12 cm
A tangent PQ is drawn at point P such that PQ = 16 cm
To Find : Length of OQ
Now, OP ⏊ PQ [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △POQ,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OQ)2 = (OP)2 + (PQ)2
(OQ2 = (12)2 + (16)2
625 = 144 + 256
(OQ)2 = 400
OQ = 20 cm
So,
Assertion is correct, and Reason is also correct.
Assertion - and - Reason Type
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A)
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
Let PT and PQ are two tangents from external point P to a circle with center O
In △OPT and △OQT
OP = OQ
[radii of same circle]
OT = OT
[common]
PT = PQ
[Tangents drawn from an external point are equal]
△OPT ≅ △OQT
[By Side - Side - Side Criterion]
∠POT = ∠QOT
[Corresponding parts of congruent triangles are congruent]
i.e. Assertion is true
Now,
Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
As ABCD is a parallelogram
AB = CD and BC = AD
[opposite sides of a parallelogram are equal] [1]
Now, As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
AB + AB = BC + BC [From 1]
AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, reason is also true, but not a correct reason for assertion.
Hence, B is correct option .
Assertion - and - Reason Type
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A)
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
For Assertion :
In the given Figure,
As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
So, assertion is not true
For Reason,
Consider two concentric circles with common center O and AB is a chord to outer circle and is tangent to inner circle P.
Now,
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
We know, that perpendicular from center to chord bisects the chord.
So, P bisects AB.
Reason is true
Hence, Assertion is false, But Reason is true.
In the given figure, O is the center of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?
A. 130°
B. 100°
C. 90°
D. 75°
In the given figure PT is a tangent to circle ∴ we have
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 50° = 90°
∠OPQ = 40°
Now, In △POQ
OP = OQ
∠PQO = ∠QPO = 40°
[Angles opposite to equal sides are equal]
Now,
∠ PQO + ∠QPO + ∠ POQ = 180° [
By angle sum property of triangle]
40° + 40° + ∠POQ = 180°
∠POQ = 100°
If the angle between two radii of a circle is 130° then the angle between the tangents at the ends of the radii is
A. 65°
B. 40°
C. 50°
D. 90°
Let us consider a circle with center O and OA and OB are two radii such that ∠AOB = 60° .
AP and BP are two intersecting tangents at point P at point A and B respectively on the circle .
To find : Angle between tangents, i.e. ∠APB
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + 130° = 360°
∠APB = 50°
If tangents PA and PB from a point P to a circle with center O are drawn so that ∠APB = 80° then ∠POA = ?
A. 40°
B. 50°
C. 80°
D. 60°
In △AOP and △BOP
AP = BP
[Tangents drawn from an external point are equal]
OP = OP [Common]
OA = OB
[Radii of same circle]
△AOP ≅ △BOP
[By Side - Side - Side criterion]
∠APO = ∠BPO
[Corresponding parts of congruent triangles are congruent]
∠APB = ∠APO + ∠BPO
80 = ∠APO + ∠APO
2∠APO = 80
∠APO = 40°
In △AOP
∠APO + ∠AOP + ∠OAP = 180°
[By angle sum property]
40° + ∠AOP + 90° = 180°
[ ∠OAP = 90° as OA ⏊ AP because Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠AOP = 50°
In the given figure, AD and AE are the tangents to a circle with center O and BC touches the circle at F. If AE = 5 cm then perimeter of ΔABC is
A. 15 cm
B. 10 cm
C. 22.5 cm
D. 20 cm
Given : From an external point A, two tangents, AD and AE are drawn to a circle with center O. At a point F on the circle tangent is drawn which intersects AE and AD at B and C, respectively. And AE = 5 cm
To Find : Perimeter of △ABC
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
BE = BF …[1]
[Tangents from point B]
CF = CD …[2]
[Tangents from point C]
Now Perimeter of Triangle abc
= AB + BC + AC
= AB + BF + CF + AC
= AB + BE + CD + AC …[From 1 and 2]
= AE + AD
Now,
AE = AD = 5 cm as tangents drawn from an external point to a circle are equal
So we have
Perimeter = AE + AD = 5 + 5 = 10 cm
In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively.
If AB = x cm, BC = 7 cm, CR = 3 cm and
AS = 5 cm, find x.
As we know, Tangents drawn from an external point are equal.
CR = CQ [tangents from point C]
CQ = 3 cm [as CR = 3 cm]
Also,
BC = BQ + CQ
7 = BQ + 3 [BC = 7 cm]
BQ = 4 cm
Now,
BP = BQ [tangents from point B]
BP = 4 cm …[1]
AP = AS [tangents from point A]
AP = 5 cm [As AC = 5 cm] ….[2]
AB = AP + BP = 5 + 4 = 9 cm [From 1 and 2]
AB = x = 9 cm
In the given figure, PA and PB are the tangents to a circle with center O. Show that the points A, 0, B, P are concyclic.
Given : PA and PB are tangents to a circle with center O
To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.
Proof :
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]
Hence Proved.
In the given figure, PA and PB are two tangents from an external point P to a circle with center O. If ∠PBA = 65°, find ∠OAB and ∠APB.
In the given Figure,
PA = PB
[Tangents drawn from an external points are equal]
∠PBA = ∠PAB
[Angles opposite to equal sides are equal]
∠PBA = ∠PAB = 65°
In △APB
∠PAB + ∠PBA + ∠APB = 180°
65° + 65° + ∠APB = 180°
∠APB = 50°
Also,
OB ⏊ AP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAP = 90°
∠OAB + ∠PAB = 90°
∠OAB + 65° = 90°
∠OAB = 25°
Two tangent segments BC and BD are drawn to a circle with center O such that ∠CBD = 120°. Prove that OB = 2BC.
Given : A circle with center O , BC and BD are two tangents such that ∠CBD = 120°
To Proof : OB = 2BC
Proof :
In △BOC and △BOD
BC = BD
[Tangents drawn from an external point are equal]
OB = OB
[Common]
OC = OD
[Radii of same circle]
△BOC ≅ △BOD [By Side - Side - Side criterion]
∠OBC = ∠OBD
[Corresponding parts of congruent triangles are congruent]
∠OBC + ∠OBD = ∠CBD
∠OBC + ∠OBC = 120°
2 ∠OBC = 120°
∠OBC = 60°
In △OBC
⇒OB = 2BC
Hence Proved !
Fill in the blanks.
(i) A line intersecting a circle in two distinct points is called a ……..
(ii) A circle can have ………..parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the ………..
(iv) A circle can have ………..tangents.
(i) secant
(ii) two
(iii) point of contact
(iv) infinitely many
Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Let us consider a circle with center O.
TP and TQ are two tangents from point T to the circle.
To Proof : PT = QT
Proof :
OP ⏊ PT and OQ ⏊ QT
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPT = ∠OQT = 90°
In △TOP and △QOT
∠OPT = ∠OQT
[Both 90°]
OP = OQ
[Common]
OT = OT
[Radii of same circle]
△TOP ≅ △QOT
[By Right Angle - Hypotenuse - Side criterion]
PT = QT
[Corresponding parts of congruent triangles are congruent]
Hence Proved.
Prove that the tangents drawn at the ends of the diameter of a circle are parallel.
Let AB be the diameter of a circle with center O.
CD and EF are two tangents at ends A and B respectively.
To Prove : CD || EF
Proof :
OA ⏊ CD and OB ⏊ EF
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAD = ∠OBE = 90°
∠OAD + ∠OBE = 90° + 90° = 180°
Considering AB as a transversal
⇒ CD || EF
[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]
In the given figure, if AB = AC, prove that BE = CE.
We know, that tangents drawn from an external point are equal.
AD = AF
[tangents from point A] [1]
BD = BE
[tangents from point B] [2]
CF = CE
[tangents from point C] [3]
Now,
AB = AC [Given] …[4]
Substracting [1] From [4]
AB - AD = AC - AF
BD = CF
BE = CE [From 2 and 3]
Hence Proved.
If two tangents are drawn to a circle from an external point, show that they subtend equal angles at the center.
Let PT and PQ are two tangents from external point P to a circle with center O
To Prove : PT and PQ subtends equal angles at center i.e. ∠POT = ∠QOT
In △OPT and △OQT
OP = OQ [radii of same circle]
OT = OT [common]
PT = PQ [Tangents drawn from an external point are equal]
△OPT ≅ △OQT [By Side - Side - Side Criterion]
∠POT = ∠QOT [Corresponding parts of congruent triangles are congruent]
Hence, Proved.
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Let us consider a circle with center O and BC be a chord, and AB and AC are tangents drawn at end of a chord
To Prove : AB and AC make equal angles with chord, i.e. ∠ABC = ∠ACB
Proof :
In △ABC
AB = PC
[Tangents drawn from an external point to a circle are equal]
∠ACB = ∠ABC
[Angles opposite to equal sides are equal]
Hence Proved.
Prove that the parallelogram circumscribing a circle, is a rhombus.
Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
To Proof : ABCD is a rhombus.
As ABCD is a parallelogram
AB = CD and BC = AD …[1]
[opposite sides of a parallelogram are equal]
Now, As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
AB + AB = BC + BC [From 1]
AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, ABCD is a rhombus.
Hence Proved.
Two concentric circles are of radii 5 cm and 3 cm respectively. Find the length of the chord of the larger circle which touches the smaller circle.
Given : Two concentric circles (say C1 and C2) with common center as O and radius r1 = 5 cm and r2 = 3 cm respectively.
To Find : Length of the chord of the larger circle which touches the circle C2. i.e. Length of AB.
As AB is tangent to circle C2 and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
We have,
(OP)2 + (PB)2 = (OB)2
r22 + (PB)2 = r12
(3)2 + (PB)2= (5)2
9 + (PB)2 = 25
(PB)2 = 16
PB = 4 cm
Now, AP = PB ,
[as perpendicular from center to chord bisects the chord and OP ⏊ AB]
So,
AB = AP + PB = PB + PB = 2PB = 2(4) = 8 cm
A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal.
Let us consider a quadrilateral ABCD, And a circle is circumscribed by ABCD
Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.
To Proof : Sum of opposite sides are equal, i.e. AB + CD = AD + BC
Proof :
In the Figure,
As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence Proved.
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Consider a quadrilateral, ABCD circumscribing a circle with center O and AB, BC, CD and AD touch the circles at point P, Q, R and S respectively.
Joined OP, OQ, OR and OS and renamed the angles (as in diagram)
To Prove : Opposite sides subtends supplementary angles at center i.e.
∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°
Proof :
In △AOP and △AOS
AP = AS
[Tangents drawn from an external point are equal]
AO = AO
[Common]
OP = OS
[Radii of same circle]
△AOP ≅ △AOS
[By Side - Side - Side Criterion]
∠AOP = ∠AOS
[Corresponding parts of congruent triangles are congruent]
∠1 = ∠2 …[1]
Similarly, We can Prove
∠3 = ∠4 ….[2]
∠5 = ∠6 ….[3]
∠7 = ∠8 ….[4]
Now,
As the angle around a point is 360°
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360° [From 1, 2, 3 and 4]
2(∠2 + ∠3 + ∠6 + ∠7) = 360°
∠AOB + ∠COD = 180°
[As, ∠2 + ∠3 = ∠AOB and ∠5 + ∠6 = ∠COD] [5]
Also,
∠AOB + ∠BOC + ∠COD + ∠AOD = 360°
[Angle around a point is 360°]
∠AOB + ∠COD + ∠BOC + ∠AOD = 360°
180° + ∠BOC + ∠AOD = 360° [From 5]
∠BOC + ∠AOD = 180°
Hence Proved
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the center.
Let us consider a circle with center O and PA and PB are two tangents to the circle from an external point P
To Prove : Angle between two tangents is supplementary to the angle subtended by the line segments joining the points of contact at center, i.e. ∠APB + ∠AOB = 180°
Proof :
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP, By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°
Hence Proved
PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure.
Find the length of TP.
Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T
To Find : Length of PT
Construction : Join OQ
Now in △OPT and △OQT
OP = OQ
[radii of same circle]
PT = PQ
[tangents drawn from an external point to a circle are equal]
OT = OT
[Common]
△OPT ≅ △OQT
[By Side - Side - Side Criterion]
∠POT = ∠OQT
[Corresponding parts of congruent triangles are congruent]
or ∠POR = ∠OQR
Now in △OPR and △OQR
OP = OQ
[radii of same circle]
OR = OR [Common]
∠POR = ∠OQR [Proved Above]
△OPR ≅ △OQT
[By Side - Angle - Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Now,
∠ORP + ∠ORQ = 180°
[Linear Pair]
∠ORP + ∠ORP = 180°
∠ORP = 90°
⇒ OR ⏊ PQ
⇒ RT ⏊ PQ
As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have
PR = QR = PQ/2 = 16/2 = 8 cm
∴ In right - angled △OPR,
By Pythagoras Theorem
[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]
(OP)2 = (OR)2 + (PR)2
(10)2 = (OR)2 + (8)2
100 = (OR)2 + 64
(OR)2= 36
OR = 6 cm
Now,
In right angled △TPR, By Pythagoras Theorem
(PT)2 = (PR)2 + (TR)2 [1]
Also, OP ⏊ OT
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In right angled △OPT, By Pythagoras Theorem
(PT)2 + (OP)2 = (OT)2
(PR)2 + (TR)2 + (OP)2= (TR + OR)2 [From 1]
(8)2 + (TR)2 + (10)2 = (TR + 6)2
64 + (TR)2 + 100 = (TR)2 + 2(6)TR + (6)2
164 = 12TR + 36
12TR = 128
TR = 10.7 cm [Appx]
Using this in [1]
PT2 = (8)2 + (10.7)2
PT2 = 64 + 114.49
PT2 = 178.49
PT = 13.67 cm [Appx]