Theorems Related To Tangent To A Circle

Theory.

⇒ Exterior angle is sum of opposite interior angles

⇒ if 2 sides of triangle are equal then their corresponding angles will also be equal

Solution.

In Δ AOB

AO = OB ∵ (radius of same circle )

∴ ∠OAB = ∠OBA = 21°

In Δ BOT

∠BOT = ∠OAB + ∠OBA ∵ (Exterior angle property)

= 21° + 21° = 42°

∠OBT = 90° ∵ (Radius of circle from point of contact of tangent is 90° )

∠OBT + ∠BOT + ∠BTO = 180° ∵ (Angle sum property)

90° + 42° + ∠BTO = 180°

∠BTO = 180° - 132° = 48°

∠BTO = ∠BTA = 48°

In the figure above, we have to prove that XA bisects angle YXZ.

Angle XAY = 90°

And XA=AY

Thus ∠AXY = AYX

In Δ AXY,

∠AXY+∠AYX+∠XAY =180° (As sum of angles of a triangle =180°

Thus,

2∠AXY=90° (∠AXY=∠AYX and ∠XAY=90°)

∠AXY=45°

∠ZXY=90°

Thus, ∠AXZ=45°

Hence, XA bisects ∠YXZ

Theory.

⇒ Angle sum property of triangle is 180°

⇒ if 2 sides of triangle are equal then their corresponding angles will also be equal

Solution.

In Δ PRS

PS = PR

∴ ∠PSR = ∠PRS

∠RPS = 90° ∵ (Radius of circle from point of contact of tangent is 90° )

∠PSR + ∠PRS + ∠RPS = 180°

2∠PSR = 180° - 90°

∠PSR = 45°

∠PSR = ∠PRS = 45°

In Δ PRT

∠PTR = 90° ∵ (3^rd point of triangle on circumference of semicircle is always 90° )

∠PRT = ∠PRS = 45°

∠TPR + ∠PRT + ∠PTR = 180°

∠TPR = 180° - 135°

= 45°

∠PRT = ∠TPR

RT = TP ∵ (isosceles triangle property)………1

In Δ PTS

∠RPS = ∠TPS + ∠TPR = 90°

∠TPS + 45° = 90°

∠TPS = 45°

∠PST = ∠PSR = 45° ∵ (proved above)

∠PST = ∠TPS

PT = ST ∵ (isosceles triangle property)………2

Joining 1 and 2

We get ;

PT = ST = RT

Formula used.

⇒ Perpendicular of tangent through point of contact pass through centre of circle

⇒ Sum of all angles of quadrilateral is 360°

⇒ Diagonals of square bisect each other at 90°

Solution.

Join AB and OT

In quadrilateral OATB

As OA and OB are perpendicular to each other

∠AOB = 90°

If tangents from point A and B are drawn

Then;

∠OAT = ∠OBT = 90°

Sum of all angles of quadrilateral is 360°

∠OAT + ∠OBT + ∠AOB + ∠ATB = 360°

90° + 90° + 90° + ∠ATB = 360°

∠ATB = 360° - 270°

= 90°

All angles of quadrilateral are 90°

Hence quadrilateral can be either square or rectangle

OA = OB ∵ (Both are radius of same circle)

If adjacent sides are equal

Then given quadrilateral is a square

In square diagonals are equal

Hence AB = OT

And diagonal bisect each other at 90°

Formula used.

⇒ Mid-point theorem.

Line joining mid-point of 2 sides of triangle is half of 3^rd side of triangle

⇒ Perpendicular to chord divides the chord in 2 equal parts

Solution

As AB and AC are tangents for smaller circle

And Chords for bigger circle

The perpendicular of tangents passes through centre

And radius of smaller circle act as perpendicular to chord

Perpendicular of chord divides it into equal parts

∴ P and Q are mid-points of AB and AC

Join BC to form Δ ABC

As P and Q are mid-points of AB and AC

By mid-point theorem

Line joining mid-point of 2 sides of triangle is half of 3^rd side

Hence;

PQ = 1/2 BC

Formula used.

⇒ Perpendicular to tangent pass through centre of circle

⇒ Mid-point of chord is perpendicular line passes through centre

Solution.

As we join the figure

If P is mid-point of chord YZ

Then;

Line passing through centre to mid-point of line is perpendicular

Therefore OP is perpendicular to YZ

∠P = 90°

As there is tangent from point A on circle

Line passing through centre and point of contact is perpendicular to tangent

∠A = 90°

In Quadrilateral XAOP

∠A + ∠P = 90° + 90° = 180°

∠A + ∠P + ∠O + ∠X = 360°

∠O + ∠X = 360° - 180° = 180°

Sum of both opposite angles are 180°

∴ Quadrilateral XAOP is cyclic quadrilateral

Formula used

⇒ Isosceles triangle property

If 2 angles of triangle are equal then their corresponding sides are also equal

⇒ Perpendicular drawn through tangent pass through centre

Solution

In Δ QPO

∠QOP = 90°

By angle sum property

∠QOP + ∠QPO + ∠OQP = 180°

∠QPO = 90° - ∠OQP

In Δ QOR

As OQ = OR ∵ radius of same circle

∠OQP = ∠ORP

As RS is tangent

∠ORS = 90°

In Δ SPR

∠SPR = ∠OPQ ∵ (Vertically opposite angles)

∠SPR = 90° - ∠OQP

∠SRP = ∠ORS - ∠ORP

= 90° - ∠OQP

Hence;

∠SPR = ∠SRP

By isosceles triangle property

∴ SR = SP

Formula used.

⇒ Isosceles triangle property

If 2 angles of triangle are equal then their corresponding sides are also equal

⇒ Perpendicular drawn through tangent pass through centre

Solution

Join OR

As QP is tangent at point Q and RP tangent to point R

Hence;

∠OQP = ∠ORP = 90°

In Δ OQR

As OQ = OR

By isosceles triangle property

∠OQR = ∠ORQ

In Δ PQR

By angle sum property

∠P + ∠PQR + ∠PRQ = 180°

∠P + [90° - ∠OQR] + [90° - ∠ORQ] = 180°

∠P + ∠OQR + ∠ORQ = 180° - 180°

∠P = ∠OQR + ∠ORQ

∠P = 2∠OQR

∠P = 2∠MQR

Formula used.

⇒ Isosceles triangle property

If 2 angles of triangle are equal then their corresponding sides are also equal ⇒ Perpendicular drawn through tangent pass through centre

Solution

Mark centre of circle to be X

Join AX, CX, BX, DX radius of circle

In Δ BDX

As XB = XD

By isosceles triangle property

∠DBX = ∠BDX

In Δ ACX

As XA = XC

By isosceles triangle property

∠ACX = ∠CAX

In quadrilateral APBO

∠OAP = ∠XAP - ∠XAC

= 90° - ∠XAC ∵ (AP is tangent at point A)

∠OBP = ∠XBP + ∠XBD

= 90° + ∠XBD ∵ (BP is tangent at point B)

By angle sum property

∠OBP + ∠P + ∠OAP + ∠AOB = 360°

∠AOB = 360° - ∠P - ∠OBP - ∠OAP

= 360° - ∠P – [90° - ∠XAC ] – [90° + ∠XBD]

= 180° - ∠P + ∠XAC – ∠XBD

In quadrilateral DQCO

∠ODQ = ∠XDQ - ∠XDB

= 90° - ∠XDB ∵ (AP is tangent at point A)

∠OCQ = ∠XCQ + ∠XCA

= 90° + ∠XCA ∵ (BP is tangent at point B)

By angle sum property

∠ODQ + ∠Q + ∠OCQ + ∠DOC = 360°

∠DOC = 360° - ∠Q - ∠ODQ - ∠OCQ

= 360° - ∠Q – [90° - ∠XDB ] – [90° + ∠XCA]

= 180° - ∠Q + ∠XDB – ∠XCA

As ∠DBX = ∠BDX and ∠ACX = ∠CAX

= 180° - ∠Q + ∠XBD – ∠XAC

∠AOB + ∠DOC

= 180° - ∠P + ∠XAC – ∠XBD + [180° - ∠Q + ∠XBD – ∠XAC]

= 360° - ∠P - ∠Q

∠BOC = ∠AOD ∵ (Vertically opposite angle)

[∠BOC + ∠AOD] + [∠AOB + ∠DOC] = 360°

2∠BOC + [360° - ∠P - ∠Q] = 360°

2∠BOC = ∠P + ∠Q

Let O be the centre, B be the external point and A be the point where the tangent meets the circle.

OA is the radius of the circle

Diameter = 16 cm

Radius = OA = 8 cm

OB = 17 cm

Δ AOB is a right angled triangle with ∠ A = 90⁰

Applying Pythagoras theorem :

AB² = AO²-OB²

⇒ AB² = 17²-8²

⇒ AB² = 289-64

⇒ AB² = 225

⇒ AB = 15 cm

The length of the tangent drawn to the circle from the external point is 15 cm

PAQ = 60

AQ = AP (Tangent drawn from an external source to the same circle are always equal in length)

So Δ PQA is isosceles triangle

∠ AQP = ∠ APQ = x (Let)

Since sum of interior angles of a triangle = 180⁰,so we can say,

2x + 60⁰ = 180⁰

⇒ 2x = 120⁰

⇒ x = 60⁰

∠ APQ = 60⁰

In Δ APO and Δ AQO

AP = AQ (Tangents drawn to the same circle from an external point are always equal in length)

AO = AO (Common)

∠ OPA = ∠ OQA (Tangents are perpendicular to the line joining the centre)

So Δ APO and Δ AQO are congruent by S.A.S. axiom of congruency.

So from corresponding parts of congruent triangle we get

∠ POA = ∠ QOA = x (Let)

∠ POA + ∠ QOA = 2x

∠ QOR = 180⁰-2x (Angle on a straight line)

∠ OQR = ∠ ORQ (Δ OQR is isosceles)

∠ OQR + ∠ ORQ = 180⁰-(180⁰-2x) = 2x

Since they are equal so

∠ OQR = x = ∠ AOQ

∠ OQR and ∠ AOQ forms a pair of alternate interior angles

So OA∥ RQ

Let us assume ∠ OAD = ∠ OAB = a

∠ OBC = ∠ OBA = b

∠ OCD = ∠ OCB = c

∠ ODC = ∠ ODA = d

Since ABCD is a quadrilateral, so

2 (a + b + c + d) = 360⁰

⇒ a + b + c + d = 180⁰ …Equation (i)

In Δ AOB

∠ AOB = 180⁰- (a + b)

In Δ COD

∠ COD = 180⁰-(c + d)

∠ AOB + ∠ COD = 360⁰–(a + b + c + d)

Putting the value from Equation (i) we get

∠ AOB + ∠ COD = 360⁰–180⁰

⇒ ∠ AOB + ∠ COD = 180⁰

Let ABCD be the parallelogram

AP and AS are tangents drawn from point A

BP and BQ are tangents drawn from point B

CR and CQ are tangents drawn from point C

DR and DS are tangents drawn from point D

Each of the above pairs of tangents are equal to each other in length since tangents drawn from an external point are always equal.

Using this concept we can say

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

Adding each length segment we can say

⇒ AB + CD = AD + BC

Since Opposite sides of a parallelogram are equal so

2AB = 2AD

⇒ AB = AD

Since adjacent sides of a parallelogram are equal, so it’s a rhombus.

∠ COD = 56⁰

∠ COE = 40⁰

∠ DAC and ∠ DOC are supplementary to each other since AD and AC are perpendicular to OD and OC respectively

⇒ ∠ DAC = (180⁰-56⁰) = 124⁰

Since DA = AC so Δ DAC is isosceles

∠ ACD = 28⁰

∠ COE and ∠ CBE are supplementary to each other since BC and BE are perpendicular to OC and OE respectively

⇒ ∠ CBE = (180⁰-40⁰) = 140⁰

Since BC = BE so Δ CBE is isosceles

∠ BCE = 20⁰

y-x = ∠ ACD-∠ BCE = 8⁰

∠ CDO = ∠ OCD = 62⁰ (∠ ADO and ∠ ACO is equal to 90⁰)

Hence OD = OC

∠ OEC = ∠ OCE = 70⁰(∠ OCB and ∠ OEB is equal to 90⁰)

Hence OE = OC

So combining the above two we can say

OD = OC = OE

Let the radius of the circle with centre A be R_a, B be R_b and O be R_o

Length OA = Radius of circle O + Radius of circle A

⇒ OA = R_o + R_a

Length OB = Radius of circle O + Radius of circle B

⇒ OB = R_o + R_b

⇒AO + BO = R_o + R_a + R_o + R_b

⇒AO + BO = 2R_o + R_a + R_b

Since the radius is always a constant quantity so AO + BO is also a constant quantity.

O is the point the two circles meet

AO and AP are the radius of circle A

BO and BQ are the radius of circle B

∠ POA = ∠ BOQ (Vertically Opposite)

∠ POA = ∠ OPA (Δ AOP is an isosceles triangle)

∠ BOQ = ∠ BQO (Δ BOQ is an isosceles triangle)

So combining the above two relations we can say

∠ OPA = ∠ BQO

Hence the above two angles forms a pair of alternate interior angle

So we say AP∥ BQ

Let three equal circles be there with centre A, B and C

Let the radius of each circle be equal to r since all the circles are equal

Since the three circles touch each other externally so the length of each side of the triangle is a sum of the radius of each circle.

So we can say each side of the triangle is equal to

AB = 2r

BC = 2r

CA = 2r

Since AB = BC = CA, so we can say that the triangle is equilateral.

AB and AC are two tangents from external point

Since DE is a variable tangent, it meets the circle at a variable point X

AB = AC (Tangents drawn from an external point to the same circle are always equal)

So we can say

2AB = AB + AC

⇒ 2AB = (AD + DB) + (AE + EC)

⇒ 2AB = (AD + AE) + (DB + EC) …Equation(i)

DB = DX (Tangents drawn from an external point to the same circle are always equal)

XE = EC (Tangents drawn from an external point to the same circle are always equal)

Replacing DB by DX and EC by XE we get

⇒ 2AB = (AD + AE) + (DX + XE)

⇒ 2AB = AD + AE + DE

⇒ 2AB = Perimeter of Δ ADE

The ΔABO formed by joining the three points is always a right angled triangle

AO is the hypotenuse of Δ ABO

Applying Pythagoras theorem :

AB² = AO²-OB²

⇒ AB² = 13²-5²

⇒ AB² = 169-25

⇒ AB² = 144

⇒ AB = 12 cm.

So Correct Option is (A)

Let D be the point where the transverse tangent meets the direct tangent

∠ DAC = ∠ DCA (Tangents drawn from an external point to the same circle are always equal and hence Δ DAC is isosceles)

Let ∠ DAC = ∠ DCA = a …Equation(i)

∠ DBC = ∠ DCB (Tangents drawn from an external point to the same circle are always equal and hence Δ DBC is isosceles)

Let ∠ DBC = ∠ DBA = b …Equation(ii)

From Δ ABC we get

∠ ACB = 180⁰-(a + b)

From Equation (i) and (ii) we get

∠ DCA + ∠ DCB = (a + b)

⇒ ∠ ACB = (a + b)

Equating ∠ ACB found in the above two cases we get

180⁰-(a + b) = (a + b)

⇒ (a + b) = 90⁰

So ∠ ACB = 90⁰

So D is the correct option

Length of tangent using Pythagoras theorem = √(13²-5²)

⇒ Length of tangent = √ 144 = 12 cm

PQ = PR = 12 cm

So there are two right angles in the quadrilateral PQOR

Area of quadrilateral PQOR = (PQ× Radius) = 60 sq cm.

So A is the correct Option.

Since the two circles touch each other externally so the distance between the two centre is always added

Distance between the two centre = (5 + 3) = 8 cm .

So the correct Option is D

Since the two circles touch each other internally so the distance between the two centre is always subtracted

Distance between the two centre = (3.5-2) = 1.5 cm.

(i) True

Tangents can only meet at a point on the circumference of a circle but cannot meet an internal point.

(ii) False

There can be only two parallel tangents that can be drawn on a circle which will lie diametrically opposite to each other.

(i) chord

If a straight line intersects the circles at two points, then the straight line is called chord of circle .

(ii) 4

If two circles do not intersect or touch each other, then the maximum number of common tangents can be drawn is 4 .

(iii) transverse

Two circles touch each other externally at the point A. A common tangent drawn to two circles at the point A is transverse common tangent.

∠ PBO = 30⁰ (Given)

∠ OPB = 30⁰ (Δ PBO is an isosceles triangle)

∠ OPT = 90⁰ (Tangents are perpendicular to the line joining the centre of the circle)

∠ BPT = (90⁰ + 30⁰) = 120⁰

∠ PTA = 180⁰–(120⁰ + 30⁰) = 30⁰ (Sum of interior angles of Δ BPT)

AP = 4 cm

BP = 6cm

AC = 12 cm

AP = AR = 4 cm (Tangents drawn from an external point to the same circle are always equal)

CR = (AC-AR) = 8 cm

CR = CQ = 8 cm (Tangents drawn from an external point to the same circle are always equal)

BP = BQ = 6 cm (Tangents drawn from an external point to the same circle are always equal)

x = BC = (BQ + CQ) = 14 cm

Let radius of circle with centre A be x, with centre B be y and with centre C be z

x + y = 5

⇒ y = 5-x …Equation(i)

y + z = 7

Putting the value of y from Equation (i) in the above equation

5-x + z = 7

⇒ z – x = 2 …Equation (ii)

z + x = 6 …Equation (iii)

Adding Equation (ii) and (iii) we get

2z = 8

⇒ z = 4 cm

Putting the value of z in Equation (iii) we get

4 + x = 6

⇒ x = 2 cm

The length of radius of circle with centre A = 2 cm

CP = CQ = 11 cm (Tangents drawn from an external point to the same circle are always equal)

BC = 7 cm (Given)

⇒ BQ = (CQ-BC) = 4 cm

BQ = BR (Tangents drawn from an external point to the same circle are always equal)

⇒ BR = 4 cm

A and B are the centre of the two circles

CD is the common tangent

Radius of circle A = 8 cm

Radius of circle B = 3 cm

AB = 13 cm

Since ED∥ AB so

ED = 13 cm

EC = (8-3) = 5cm

Applying Pythagoras theorem in Δ CED we get

Length of tangent = √(13²-5²)

⇒ Length of tangent = √ 144 = 12 cm