Circles

Given : Two circles (say C1 and C2) with common center as O and

Radius of circle C1, r₁ = 4 cm

Radius of circle C2, r₂ = 5 cm

Also say AC is the chord of circle C2 which is tangent to circle C1 and OB is the radius of circle to the point of contact of tangent AC .

To find : Length of chord AC

Now, Clearly OB ⏊ AC [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

So OBC is a right-angled triangle

So, it will satisfy Pythagoras theorem [ i.e. (base)² + (perpendicular)² = (hypotenuse)² ]

i.e.

(OB)² + (BC)² = (OC)²

As OB and OC are the radii of circle C1 and C2 respectively.

So

(4)² + (BC)² = (5)²

16 + (BC)² = 25

(BC)² = 25 - 16 = 9

BC = 3 cm

Also

AB = BC [ Perpendicular through the center to a chord in a circle (C2 in this case) bisects the chord]

And

AC = AB + BC

= AB + AB = 2AB = 2(3) = 6 cm

As in the given figure ABCD is a quadrilateral circumscribing the circle and we know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

So, we have

∠AOB + ∠COD = 180°

125° + ∠COD = 180°

∠COD = 55°

Given : A circle with center O and AC is its diameter and ∠ACB = 50° , Also AT is a tangent to the circle at point A
To Find : ∠BAT

We Have

∠CBA = 90° [ As angle in a semicircle is a right angle]

So, By angle sum property of a triangle

∠ACB + ∠CAB + ∠CBA = 180°
50° + ∠CAB + 90° = 180°

∠CAB = 40° [1]

Also

OA ⏊ AT [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

So, ∠OAT = 90°

∠OAT + ∠BAT = 90°

∠CAT + ∠BAT = 90°

40° + ∠BAT = 90° [ using 1]

∠BAT = 50°

Let us draw a circle of radius 5 cm having center O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.

Also, OQ = OR = radius = 5cm [1]

And OP = 13 cm

As OQ ⏊ PQ and OR ⏊PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

△POQ and △POR are right-angled triangles.

In △PQO By Pythagoras Theorem [ i.e. (base)²+ (perpendicular)²= (hypotenuse)² ]

(PQ)² + (OQ)²= (OP)²

(PQ)² + (5)² = (13)²

(PQ)² + 25 = 169

(PQ)² = 144

PQ = 12 cm

And

PQ = PR = 12 cm [ tangents through an external point to a circle are equal] [2]

Area of quadrilateral PQRS, A = area of △POQ + area of △POR

[Using 1 and 2]

Let's make a diagram for the given problem in which we have a circle having AB as diameter and O as center and having radius 5cm, and a tangent XAY at point A on the circle .

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.

∠BAX + ∠AEC = 180° [Sum of co-interior angles is 180°]

Now, ∠OAX =∠BAX= 90°

[Tangent and any point of a circle is perpendicular to the radius through the point of contact]

So we have

90 + ∠AEC = 180

∠AEC = 90°

Therefore, ∠OEC = 90°

So, OE ⏊ CD and △OCB is a right-angled triangle.

Now in △OEB

By Pythagoras theorem [ i.e. (base)² + (perpendicular)² = (hypotenuse)² ]

(OE)² + (EC)² = (OC)²

And we have

OC = 5 cm [radius]

OE = AE - AO = 8 - 5 = 3 cm

(3)² + (EC)²= (5)²

9 + (EC)² = 25

(EC)²= 25 - 9 = 16

EC = 4 cm

Also, CE = ED

[since, perpendicular from center to the chord bisects the chord]

So, CD = CE + ED = 2CE = 2(4) = 8 cm

Given : A circle with center O and a tangent AT passing through point A on the circle, also OT = 4 cm and ∠OTA = 30°

To find : Length of AT

Construction : Join OA.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

So we have

OA ⏊ AT

Therefore, ∠OAT = 90°

In △ OAT

Using

We have,

Given : OP is a radius and PR is a tangent in a circle with center O with ∠RPQ = 50°

To find : ∠POQ

Now, OP ⏊ PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

∠OPR = 90°

∠OPQ + ∠RPQ = 90°

∠OPQ + 50° = 90°

∠OPQ = 40°

In △POQ

OP = OQ [radii of same circle]

∠OQP = ∠OPQ = 40° [ angles opposite to equal sides are equal] [1]

In △OPQ By angle sum property of a triangle

∠OPQ + ∠OPQ + ∠POQ = 180°

40° + 40° + ∠POQ = 180°

∠POQ = 100°

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and ∠APB = 50°

To find : ∠OAB

OA ⏊ AP and OB ⏊ PB [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90° [1]

In Quadrilateral AOBP [ By angle sum property of quadrilateral]

∠OBP + ∠OAP + ∠AOB + ∠APB = 360°

90° + 90° + ∠AOB + 50° = 360°

∠AOB = 130° [2]

Now in △OAB

OA = OB [Radii of same circle]

∠OBA = ∠ OAB [3]

Also, By angle sum property of triangle

∠OBA + ∠OAB + ∠AOB = 180°

∠OAB + ∠OAB + 130 = 180 [using 2 and 3]

2∠OAB = 50°

∠OAB = 25°

Given : A circle with center O and PA and PB are two tangents to the circle at point A and C from an external point P such that ∠APC = 60° [i.e. angle of inclination between two tangents] .

To Find : AP and PC

In △OAP and △OCP

AO = OC [ radii of same circle]

OP = OP [ common ]

AP = PC [ tangents through an external point to a circle are equal]

△OAP ≅ △OPC [ By Side Side Criterion]

∠APO = ∠OPC [Corresponding parts of congruent triangles are equal] [1]

Now, ∠APC = 60° [Given]

∠APO + ∠OPC = 60°

∠APO + ∠APO = 60° [By 1]

2∠AP0 = 60°

∠APO = 30°

Now, OA ⏊ AP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OAP = 90°

So △AOP is a right-angled triangle

And we know that,

So,

[As OA is radius and equal to 3 cm]

Given : PQR is a tangent to a circle Q with center O and AB is a chord parallel to PR and ∠BQR = 70°

To Find : ∠AQB

Now, ∠OQR = ∠DQR = 90° [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

∠DQR = ∠QBD + ∠BQR

90 = ∠QBD + 70°

∠QBD = 20° [1]

As, AB || PR

∠BDQ + ∠DQR = 180° [ some of co interior angles ]

∠BDQ + 90° = 180°

∠BDQ = 90°

And

∠ADQ + ∠BDQ = 180° [Linear pair]

∠ADQ + 90° = 180°

∠ADQ = 90° = ∠BDQ

In △AQD and △BQD

QD = QD [common]

∠ADQ = ∠BDQ [Proved Above]

AD = BD [since, perpendicular from center to the chord bisects the chord]

△AQD ≅ △BQD [By Side Angle Side Criterion]

And

∠AQD = ∠BQD [ Corresponding parts of congruent triangles are equal ]

∠AQD + ∠BQD = ∠AQB

∠AQB = 2∠BQD [As ∠AQD = ∠BQD ]

∠AQB = 2(20) = 40° [From 1]

False

Consider the problem in above diagram. In which we have a circle with center O and AB be any chord with ∠AOB = 60°

Now,

OA ⏊ AC and OB ⏊ CB [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

∠OBC = ∠OAC = 90° [1]

In Quadrilateral AOBC [ By angle sum property of quadrilateral]

∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°

90° + 90° + 60° + ∠ACB = 360°

∠ACB = 120° [2]

So the angle between two tangents is 120°. So the above statement is false .

False

Length of tangent may or may not be greater than the radius of circle.

True

Consider a figure for the problem in which we have a circle with center O.

PT is a tangent drawn from external point P. Joint OT.

OT ⏊ PT [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

So, OPT is a right-angled triangle formed.

In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle.

So,

OP > PT or PT < OP

Hence, Length of tangent from an external point P on a circle with center O is always less than OP .

True

This may be possible only when both tangent lines coincide or are parallel to each other.

True

Let us consider a circle with center O and tangents PT and PR and angle between them is 90° and radius of circle is a

To show :

Proof :

In △OTP and △ORP

TO = OR [ radii of same circle]

OP = OP [ common ]

TP = PR [ tangents through an external point to a circle are equal]

△OTP ≅ △ORP [ By Side Side Side Criterion ]

∠TPO = ∠OPR [Corresponding parts of congruent triangles are equal ] [1]

Now, ∠TPR = 90° [Given]

∠TPO + ∠OPR = 90°

∠TPO + ∠TPO = 90° [By 1]

∠TP0 = 45°

Now, OT ⏊ TP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OTP = 90°

So △POT is a right-angled triangle

And we know that,

So,

[As OT is radius and equal to a]

Hence Proved !

False

Let us consider a circle with center O and tangents PT and PR and angle between them is 60° and radius of circle is a .

In △OTP and △ORP

TO = OR [ radii of same circle]

OP = OP [ common ]

TP = PR [ tangents through an external point to a circle are equal]

△OTP ≅ △ORP [ By Side Side Side Criterion ]

∠TPO = ∠OPR [Corresponding parts of congruent triangles are equal ] [1]

Now, ∠TPR = 60° [Given]

∠TPO + ∠OPR = 60°

∠TPO + ∠TPO = 60° [By 1]

∠TP0 = 30°

Now, OT ⏊ TP [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OTP = 90°

So △POT is a right-angled triangle

And we know that,

So,

[As OT is radius and equal to a]

So the above statement is false .

True

Let us consider a circle in which EF is a tangent passing through point A on the circle and ABC is an isosceles triangle in the circle , in which AB = AC

To Prove : EF || BC

Construction : Join OA , OB and OC

Proof :

AB = AC [Given]

∠ACB = ∠ABC [Angles opposite to equal sides are equal] [1]

∠EAB = ∠ACB [Tangents drawn from an external point to a circle are equal] [2]

From [1] and [2]

∠EAB = ∠ACB

i.e. EF || BC [ two lines are parallel if their alternate interior angles are equal]

False

Let the S1, S2, S3, …., Sn be n circles with centers C1, C2, C3, …, Cn respectively.

And The PQ is a common tangent to all the circles at point A which is common to all circles.

As we know,

tangent at any point on the circle is perpendicular to the radius through point of contact

we have,

C1A ⏊ PQ

C2A ⏊ PQ

C3A ⏊ PQ

CnA ⏊ PQ

So, C1 C2 C3 … Cn all lie on the perpendicular line to PQ but not on perpendicular bisector as

PA may or may not be equal to AQ .

True

Let as draw a line segment PQ and S1 S2 S3 …. Sn are n circles with center C1 C2 C3 … Cn respectively passing through p and Q

Here PQ either will be a chord to a circle or the diameter (possible only once when diameter length is equal to the length of PQ.

Now, if PQ is diameter then clearly center lies on the half of diameter i.e. on the perpendicular bisector of PQ.

But if PQ is a chord then we know that perpendicular from the center to the chord of circle bisect the chord implies that center lies on the perpendicular bisector of the chord.

So for each circle center lies on the perpendicular bisector of PQ.

So above statement is true .

True

Given : AB is a diameter of circle with center O and AC is a chord such that ∠BAC = 30° Also tangent at C intersects AB extends at D.

To prove : BC = BD

Proof :

OA = OC [radii of same circle]

∠OCA = ∠OAC = 30° [Angles opposite to equal sides are equal]

∠ACB = 90° [angle in a semicircle is a right angle.]

∠OCA + ∠OCB = 90°

30° + ∠OCB = 90°

∠OCB = 60° [1]

OC = OB [ radii of same circle]

∠OBC = ∠OCB = 60° [angles opposite to equal sides are equal]

Now, ∠OBC + ∠CBD = 180° [linear pair]

60 + ∠CBD = 180°

So, ∠CBD = 120° [2]

Also,

OC ⏊ CD [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠OCD = 90°

∠OCB + ∠BCD = 90°

60 + ∠BCD = 90

∠BCD = 30° [3]

In △BCD

∠CBD + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]

120° + 30° + ∠BDC = 180° [From 2 and 3]

∠BDC = 30° [4]

From [3] and [4]

∠BCD = ∠BDC = 30°

BC = BD [Sides opposite to equal angles are equal]

Hence Proved !

Let C₁ and C₂ are two concentric circles with center O and radius of outer circle is 5 cm.

Given : AC is a chord with length 8 cm that is tangent to inner circle .

To find : Radius of inner circle i.e. OD

AC is a tangent for C₁ at D so,

OD ⏊ AC [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

So, OAD is a right-angled triangle at D .

Also it implies that OD is perpendicular to the chord AC in C₂

So we have,

AD = DC [perpendicular from the center to the chord bisects the chord]

AC = AD + DC

8 = AD + AD
AD = 4 cm

In △OAD By Pythagoras Theorem

(OA)² = (OD)² + (AD)²

(5)² = (OD)²+ (4)²

25 = (OD)² + 16

(OD)²= 9

OD = 3 cm

Given : PQ and PR are two tangents drawn at points Q and R are drawn from an external point P .

To Prove : QORP is a cyclic Quadrilateral .

Proof :

OR ⏊ PR and OQ ⏊PQ [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠ORP = 90°

∠OQP = 90°

∠ORP + ∠OQP = 180°

Hence QOPR is a cyclic quadrilateral. As the sum of the opposite pairs of angle is 180°

Let PR and PQ are two intersecting lines [intersects on point P] touching the circle with center O and we joined OR, OQ and OP.

To Prove : Center O lies on the angle bisector of PR and PQ i.e. OP is the angle bisector of ∠RPQ

Proof :

Clearly, PQ and PQ are tangents to the circle with a common external point P.

In △POR and △POQ

OR = OQ [radii of same circle]

OP =OP [Common]

PR = PQ [Tangents drawn from an external point to a circle are equal ]

△POR ≅ △POQ [ By Side Side Side criterion ]

∠RPO = ∠OPQ [ Corresponding parts of congruent triangles are equal ]

This implies that OP is the angle bisector of ∠RPQ .

Hence Proved .

Given : BC and BD are two tangents drawn from a common external point to a circle with center O such that ∠DBC = 120°

To Prove : BC + BD = BO i.e. BO = 2BC

Proof :

In △OBC and △ODB

OC = OC [ Radii of same circle ]

OB = OB [Common]

BC = BD [Tangents drawn from an external point to a circle are equal ] [1]

△OBC ≅ △OBD [ By Side Side Criterion ]

∠CBO = ∠DBO [Corresponding parts of congruent triangles are equal ]

Also,

∠DBC = 120°

∠CBO + ∠DBO = 120°

∠CBO + ∠CBO = 120°

∠CBO = ∠DBO = 60°

OC ⏊BC [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

Therefore,

OBC is a right-angled triangle at C

So we have,

BO = 2BC

From [1] we have

BC = BD

BC + BC = BC + BD

2BC = BC + BD

BO = BC + BD

Hence Proved .

Given : AB and CD are two common tangents to two circles of unequal radii.

To Prove : AB = CD

Construction: Produce AB and CD, to intersect at P.

Proof:

Consider the circle with greater radius.

AP = CP [Tangents drawn from an external point to a circle are equal] [1]

Also,

Consider the circle with smaller radius.

BP = BD [Tangents drawn from an external point to a circle are equal] [2]

Substract [2] from [1]. We Get

AP - BP = CP - BD

AB = CD

Hence Proved .

Given: AB and CD are two common tangents to two circles of equal radii .

To Prove: AB = CD

Construction: Join OA, OC, O’B and O’D

Proof:

Now, ∠OAB = 90° and ∠OCD = 90° as OA ⏊ AB and OC ⏊ CD

[tangent at any point of a circle is perpendicular to radius through the point of contact]

Thus, AC is a straight line.

Also,

∠O'BA = ∠O'DC = 90° [Tangent at a point on the circle is perpendicular to the radius through point of contact]

Thus, BD is Also a straight line.

So ABCD is a quadrilateral with Four sides as AB, BC, CD and AD

But as

∠A = ∠B = ∠C = ∠D = 90°

So, ABCD is a rectangle.

Hence, AB = CD [opposite sides of rectangle are equal]

Given : AB and CD are two tangents to two circles which intersects at E .

To Prove : AB = CD

Proof :

As

AE = CE [Tangents drawn from an external point to a circle are equal] [1]

And

EB = ED [Tangents drawn from an external point to a circle are equal] [2]

Adding [1] and [2]

AE + EB = CE + ED

AB = CD

Hence Proved !

PQ is a chord in a circle with center O and MN is a tangent drawn at point R on the circle PQ is parallel to MN

To Prove : R bisects the arc PRQ i.e. arc PR = arc PQ

Proof :

∠1 = ∠2 [Alternate Interior angles]

∠1 = ∠3 [angle between tangent and chord is equal to angle made by chord in alternate segment]

So, we have

∠2 = ∠3

QR = PR [Sides opposite to equal angles are equal]

As the equal chords cuts equal arcs in a circle.

Arc PR = arc RQ

R bisects the arc PRQ .

Hence Proved

Let QR be a chord in a circle with center O and ∠1 and ∠2 are the angles made by tangent at point R and Q with chord respectively.

To Prove : ∠1 = ∠2

Let P be another point on the circle, then, join PQ and PR.

Since, at point Q, there is a tangent.

∠RPQ = ∠2 [angles in alternate segments are equal] [Eqn 1]

Since, at point R, there is a tangent.

∠RPQ = ∠1 [angles in alternate segments are equal] [Eqn 2]

From Eqn 1 and Eqn 2

∠1 = ∠2

Hence Proved .

Let AB be diameter in a circle with center O and MN is the chord at point A .

And CD be any chord parallel to MN intersecting AB at E

To Prove : AB bisects CD .

Proof :

OA ⏊ MN [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠MAO = 90°

Also, ∠CEO = 90°

[ if E point lies on the OA then by Corresponding angles and if E lies on OB then by sum of co-interior angles ]

So, we have OE ⏊ CD

CE = ED [ Perpendicular drawn from the center of a circle to chord bisects the chord ]

Implies that AB bisects CD

Hence Proved !

Given : A Hexagon ABCDEF circumscribe a circle .

To prove : AB + CD + EF = BC + DE + FA

Proof :

As we know, that

Tangents drawn from an external point to a circle are equal.

We have

AM = RA [1] [tangents from point A]

BM = BN [2] [tangents from point B]

CO = NC [3] [tangents from point C]

OD = DP [4] [tangents from point D]

EQ = PE [5] [tangents from point E]

QF = FR [6] [tangents from point F]

Add [1] [2] [3] [4] [5] and [6]

AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR

Rearranging

(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)

AB + CD + EF = BC + DE + FA

Hence Proved !

Given : A triangle ABC with BC = a , CA = b and AB = c . Also, a circle is inscribed which touches the sides BC, CA and AB at D, E and F respectively and s is semi- perimeter of the triangle

To Prove : BD = s - b

Proof :

Given that

Semi Perimeter = s

Perimeter = 2s

Implies that

2s = AB + BC + AC [1]

As we know,

Tangents drawn from an external point to a circle are equal

So we have

AF = AE [2] [Tangents from point A]

BF = BD [3] [Tangents From point B]

CD = CE [4] [Tangents From point C]

Adding [2] [3] and [4]

AF + BF + CD = AE + BD + CE

AB + CD = AC + BD

Adding BD both side

AB + CD + BD = AC + BD + BD

AB + BC - AC = 2BD

AB + BC + AC - AC - AC = 2BD

2s - 2AC = 2BD [From 1]

2BD = 2s - 2b [as AC = b]

BD = s - b

Hence Proved !

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 = 10 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 = 10 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = PA + PB = 10 + 10 = 20 cm

Given : A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at point A

To Prove : ∠BAT = ∠ACB

Proof :

∠ABC = 90° [Angle in a semicircle is a right angle]

In △ABC By angle sum property of triangle

∠ABC + ∠ BAC + ∠ACB = 180 °

∠ACB + 90° = 180° - ∠BAC

∠ACB = 90 - ∠BAC [1]

Now,

OA ⏊ AT [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠OAT = ∠CAT = 90°

∠BAC + ∠BAT = 90°

∠BAT = 90° - ∠BAC [2]

From [1] and [2]

∠BAT = ∠ACB [Proved]

Given : Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common chord.

To Find : Length of common chord PQ

∠OPO' = 90° [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

So OPO is a right-angled triangle at P

Using Pythagoras in △ OPO', we have

(OO')²= (O'P)²+ (OP)²

(OO')² = (4)² + (3)²

(OO')² = 25

OO' = 5 cm

Let ON = x cm and NO' = 5 - x cm

In right angled triangle ONP

(ON)²+ (PN)²= (OP)²

x² + (PN)² = (3)²

(PN)²= 9 - x² [1]

In right angled triangle O'NP

(O'N)² + (PN)²= (O'P)²

(5 - x)² + (PN)² = (4)²

25 - 10x + x² + (PN)² = 16

(PN)² = -x²+ 10x - 9[2]

From [1] and [2]

9 - x² = -x² + 10x - 9

10x = 18

x = 1.8

From (1) we have

(PN)² = 9 - (1.8)²

=9 - 3.24 = 5.76

PN = 2.4 cm

PQ = 2PN = 2(2.4) = 4.8 cm

Given : In a right angle ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Also PQ is a tangent at P

To Prove : PQ bisects BC i.e. BQ = QC

Proof :

∠APB = 90° [Angle in a semicircle is a right-angle]

∠BPC = 90° [Linear Pair ]

∠3 + ∠4 = 90 [1]

Now, ∠ABC = 90°

So in △ABC

∠ABC + ∠BAC + ∠ACB = 180°

90 + ∠1 + ∠5 = 180

∠1 + ∠5 = 90 [2]

Now ,

∠ 1 = ∠ 3[angle between tangent and the chord equals angle made by the chord in alternate segment]

Using this in [2] we have

∠3 + ∠5 = 90 [3]

From [1] and [3] we have

∠3 + ∠4 = ∠3 + ∠5

∠4 = ∠5

QC = PQ [Sides opposite to equal angles are equal]

But Also PQ = BQ [Tangents drawn from an external point to a circle are equal]

So, BQ = QC

i.e. PQ bisects BC .

Given : Tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ.

To Find : ∠RQS

PQ = PR [Tangents drawn from an external point to a circle are equal]

∠PRQ = ∠PQR [Angles opposite to equal sides are equal] [1]

In △PQR

∠PRQ + ∠PQR + ∠QPR = 180°

∠PQR + ∠PQR + ∠QPR = 180° [Using 1]

2∠PQR + ∠RPQ = 180°

2∠PQR + 30 = 180

2∠PQR = 150

∠PQR = 75°

∠QRS = ∠PQR = 75° [Alternate interior angles]

∠QSR = ∠PQR = 75° [angle between tangent and the chord equals angle made by the chord in alternate segment]

Now In △RQS

∠RQS + ∠QRS + ∠QSR = 180

∠RQS + 75 + 75 = 180

∠RQS = 30°

Given : AB is a diameter of a circle, AB is a diameter and AC is a chord Also ∠BAC = 30° .The tangent at C intersects extended AB at a point D

To Prove : BC = BD

Proof :

∠CAB = ∠BCD =30° [1] [angle between tangent and chord is equal to angle made by chord in alternate segment]

Now, ∠ACB = 90° [Angle in a semicircle is a right angle]

∠ACD = ∠ACB + ∠BCD = 90 + 30 = 120°

In triangle ACD, By Angle Sum Property

∠ACD + ∠CAD + ∠ADC = 180°

120 + ∠CAB + ∠BDC = 180°

120 + 30 + ∠BDC = 180

∠BDC = 30° [2]

From [1] and [2]
∠BCD = ∠BDC = 30°

BD = BC [Angles opposite to equal sides are equal]

Hence Proved !

Let us draw a circle in which AMB is an arc and M is the mid-point of the arc AMB. Joined AM and MB. Also TT' is a tangent at point M on the circle.

To Prove : AB || TT'

Proof :

As M is the mid point of Arc AMB

Arc AM = Arc MB

AM = MB [As equal chords cuts equal arcs]

∠ABM = ∠BAM [Angles opposite to equal sides are equal] [1]

Now,

∠BMT' = ∠BAM [angle between tangent and the chord equals angle made by the chord in alternate segment] [2]

From [1] and [2]

∠ABM = ∠BMT'

So, AB || TT' [two lines are parallel if the interior alternate angles are equal]

Hence Proved !

Given : AB and CD are two tangents with centers O and O' intersect at E .

To Prove : O, E and O' are collinear.

Construction : Join AO, OC O'D and O'B

In △AOE and △EOC

OA = OC [radii of same circle]

OE = OE [common]

AE = EC [Tangents drawn from an external point to a circle are equal]

△AOE ≅ △EOC [By Side Side Side Criterion]

∠AEO = ∠CEO [Corresponding parts of congruent triangles are equal ]

∠AEC = ∠AEO + ∠CEO = ∠AEO + ∠AEO = 2∠AEO [1]

Now As CD is a straight line

∠AED + ∠AEC = 180° [linear pair]

2∠AEO = 180 - ∠AED [From 1]

[2]

Now, In △O'ED and △O'EB

O'B = O'D [radii of same circle]

O'E = O'E [common]

EB = ED [Tangents drawn from an external point to a circle are equal]

△O'ED ≅ △O'EB [By Side Side Side Criterion]

∠O'EB = ∠O'ED [Corresponding parts of congruent triangles are equal ]

∠DEB = ∠O'EB + ∠O'ED = ∠O'ED + ∠O'ED = 2∠O'ED [3]

Now as AB is a straight line

∠AED + ∠DEB = 180 [Linear Pair]

2∠O'ED = 180 - ∠AED [From 3]

[4]

Now,

So O, E and O' lies on same line [By the converse of linear pair]

Hence Proved.

Given : A circle with center O and radius = 5cm T is a point, OT = 13 cm. OT intersects the circle at E and AB is the tangent to the circle at E .

To Find : Length of AB

OP ⏊ PT [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

By phythagoras theorem in △OPT right angled at P

(OT)² = (OP)²+ (PT)²

(13)² = (5)² + (PT)²

(PT)²= 169 - 25 = 144

PT = 12 cm

PT = TQ = 12 cm [Tangents drawn from an external point to a circle are equal]

Now,

OT = OE + ET

ET = OT - OE = 13 - 5 = 8 cm

Now, as Tangents drawn from an external point to a circle are equal .

AE = PA [1]

EB = BQ [2]

Also OE ⏊ AB [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠AEO = 90°

∠AEO + ∠AET = 180° [By linear Pair]

∠AET = 90°

In △AET By Pythagoras Theorem

(AT)² = (AE)² + (ET)²

[Here AE = PA as tangents drawn from an external point to a circle are equal]

(PT - PA)² = (PA)² + (ET)²

(12 - PA)² = (PA)² + (8)² [from 1]

144 + (PA)² - 24PA = (PA)² + 64

24PA = 80

[3]

∠AET + ∠BET = 180 [Linear Pair]

90 + ∠BET = 180

∠BET = 90

In △BET, By Pythagoras Theorem

(BT)² = (BE)²+ (ET)²

(TQ - BQ)² = (BQ)² + (ET)² [from 2]

(12 - BQ)² = (BQ)² + (8)²

144 + (BQ)² - 24BQ = (BQ)² + 64

24BQ = 80

[4]

So,

AB = AE + BE

AB = PA + BQ [From 1 and 2]

[From 3 and 4]

Given : A circle with center O in which PC is a tangent a point C and AB is a diameter which is extended to P and ∠PCA = 110°

To Find : ∠CBA

∠ACB = 90° [Angle in a semicircle is a right angle] [1]

Also,

∠PCA = ∠ACB + ∠PCB

110 = 90 + ∠PCB

∠PCB = 20°

Now, ∠PCB = ∠BAC [angle between tangent and the chord equals angle made by the chord in alternate segment]

∠BAC = 20° [2]

Now In △ABC By angle sum property of Triangle.

∠CBA + ∠BAC + ∠ACB = 180

∠CBA + 20 + 90 = 180

∠CBA = 70°

Given : In a circle, ΔABC is inscribed in which AB = AC = 6 cm and radius of circle is 9 cm

To Find : Area of triangle ABC

Construction : Join OB and OC

In △AOB and △AOC

OB = OC [Radii of same circle]

AO = AO [Common]

AB = AC = 6 cm [Given]

△AOB ≅ △AOC [By Side Side Side Criterion]

∠OAB = ∠OBC [Corresponding parts of congruent triangles are equal ]

Implies ∠MAB = ∠MBC

Now in △ABM and △AMC

AB = AC = 6 cm [Given]

AM = AM [Common]

∠MAB = ∠MBC [Proved Above]

△ABM ≅ △AMC [By Side Angle Side Criterion]

∠AMB = ∠AMC [Corresponding parts of congruent triangles are equal ]

Now,

∠AMB + ∠ AMC = 180°

∠AMB + ∠AMB = 180°

2 ∠AMB = 180

∠AMB = 90°

We know that a perpendicular from center of circle bisects the chord.

So, OA is perpendicular bisector of BC.

Let OM = x

Then, AM = OA - OM = 9 - x

[ As OA = radius = 9 cm]

In right angled ΔAMC, By Pythagoras theorem

(AM)² + (MC)² = (AC)²

(9 - x)² + (MC)² = (6)²

81 + x²- 18x + (MC)² = 36

(MC)² = 18x - x² - 45 [1]

In △OMC , By Pythagoras Theorem

(MC)² + (OM)² = (OC)²

18x - x² - 45 + (x)² = (9)²

18x - 45 = 81

18x = 126

x = 7

AM = 9 - x = 9 - 7 = 2 cm

In △AMC, By Pythagoras Theorem

(AM)² + (MC)² = (AC)²

(2)²+ (MC)² = (6)²

4 + (MC)²= 36

(MC)² = 32

MC = 4√2 cm

Now,

As MC = BM

BC = 2MC = 2(4√2) = 8√2 cm

And

Given : Two tangents are drawn from an external point A to the circle with center O, Tangent BC is drawn at a point R.

Radius of circle equals 5cm and OA = 13 cm

OA = 13 cm

To Find : Perimeter of ΔABC.

∠OPA = 90°

[tangent at any point of a circle is perpendicular to the radius through the point of contact]

(OA)² = (OP)² + (PA)²

(13)² = (5)² + (PA)²

169 - 25 = (PA)²

(PA)²= 144

PA = 12 cm

As we know that, Tangents drawn from an external point to a circle are equal.

So we have

PB = BR [1] [Tangents from point B]

CR = QC [2] [Tangents from point C]

Now Perimeter of Triangle PCD

= AB + BC + CA

= AB + BR + CR + CA

= AB + PB + QC + CA [From 1 and 2]

= PA + QA

Now,

PA = QA = 12 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = PA + QA = 12 + 12 = 24 cm