Learn Exercise 10.2 with Free Lessons & Tips

OA = 5 cm. OP = 3 cm

OP ⊥ AB 

Therefore AP^2.   + OP ^2. = OA ²

^{AP = √ 25 - 9 = √ 16 = 4 cm}

^{AB = 4 x 2 = 8 cm}

tangents drawn from an external point to a circle are equal in length

Hence AP = AS

BP= BQ

DR = DS

CR = CQ

adding we get

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

= AB + CD = AD + BC

Let us join point O to C.

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ΔOCA (SSS congruence criterion)

Therefore, P ↔ C, A ↔ A, O ↔ O

∠POA = ∠COA … (i)

Similarly, ΔOQB ΔOCB

∠QOB = ∠COB … (ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º

From equations (i) and (ii), it can be observed that

2∠COA + 2 ∠COB = 180º

∠COA + ∠COB = 90º

∠AOB = 90°

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 90°

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠OAP +∠APB+∠PBO +∠BOA = 360º

90º + ∠APB + 90º + ∠BOA = 360º

∠APB + ∠BOA = 180º

Hence, it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Given: ABCD  is a parallelogram circumscribing a circle with centre O

To prove: ABCD is a rhombus

We know tangents drawn to a circle from exterior points are equal in length

Therefore AP = AS . BP= BQ  CR = CQ DR= DS

adding the above equations :

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

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

= AB + CD  = AD + BC......(1)

Since ABCD is a parallelogram AB = DC and  AD = BC

therefore we can write (1) as :

2AB = 2 BC.  HENCE AB = BC

therefore we have :  AB = BC = DC = AD

Hence ABCD  is a rhombus

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In ABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = x (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x

Area of = 

Area of ΔOBC = 

Area of ΔOCA =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Area of ΔOAB =

Either x+14 = 0 or x − 7 =0

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Consider ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus, ∠POA = ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

(∠1 + ∠2) + (∠5 + ∠6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that ∠BOC + ∠DOA = 180º

Hence, it is proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior angles = 360

∠OAP + ∠APB +∠PBO + ∠BOA = 360

90 + 80 +90º +BOA = 360

∠BOA = 100

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

Hence,option (A) is correct.

Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

OP² + PQ² = OQ²

OP² + 24² = 25²

OP² = 625 − 576

OP² = 49

OP = 7

Therefore, the radius of the circle is 7 cm.

Hence, option (A) is correct.

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90º

∠OQT = 90º

In quadrilateral POQT,

Sum of all interior angles = 360

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360

⇒ 90+ 110º + 90 +PTQ = 360

⇒ PTQ = 70

Hence, option (B) is correct.

Let us consider a circle centered at point O.

AB is a tangent drawn on this circle from point A.

Given that,

OA = 5cm and AB = 4 cm

In ΔABO,

OB ⊥ AB (Radius ⊥ tangent at the point of contact)

Applying Pythagoras theorem in ΔABO, we obtain

AB² + BO² = OA²

4² + BO² = 5²

16 + BO² = 25

BO² = 9

BO = 3

Hence, the radius is 3 cm.

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.