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Learn Exercise 10.2 with Free Lessons & Tips

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

OA = 5 cm. OP = 3 cm

OP ⊥ AB 

Therefore AP2.   + OP 2. = OA 2

AP = √ 25 - 9 = √ 16 = 4 cm

AB = 4 x 2 = 8 cm

Comments

A quadrilateral ABCD is drawn to circumscribe a circle (see the figure). Prove that AB + CD = AD + BC

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

Comments

In the figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that .

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°

Comments

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-segment joining the points of contact at the centre.

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.

Comments

Prove that the parallelogram circumscribing a circle is a rhombus.

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

Comments

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see the figure). Find the sides AB and AC.

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, = −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

Comments

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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.

Comments

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of , then  is equal to:

A) B) C) D)

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.

Comments

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

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 + PQ= OQ2

OP+ 24= 252

OP= 625 − 576

OP= 49

OP = 7

Therefore, the radius of the circle is 7 cm.

Hence, option (A) is correct.

 

Comments

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that , then  is equal to:

(A)   (B)   (C)   (D)

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.

Comments

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

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

4+ BO2 = 52

16 + BO2 = 25

BO2 = 9

BO = 3

Hence, the radius is 3 cm.

Comments

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

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.

Comments

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,

∠O’PB = 90° … (1)

O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠OPB = 90° … (2)

Comparing equations (1) and (2), we obtain

∠O’PB = ∠OPB … (3)

From the figure, it can be observed that,

∠O’PB < ∠OPB … (4)

Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

Comments

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