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CIRCLES

A. SUMMATIVE ASSESSMENT

10.1 TANGENT TO A CIRCLE

1. A circle may be regarded as a collection ofpoints in a plane at a fixed distance from a fixedpoint. The fixed point is called the centre of thecircle. The fixed distance between the centre ofthe circle and the circumference, is calledradius.

2. The perimeter of the circle is referred to asthe circumference of the circle.

3. A chord of a circle is a line segment joiningany two points on the circumference.

4. An arc of a circle is a part of thecircumference.

5. A diameter of a circle is a chord whichpasses through the centre of the circle.

6. A line, which intersects the circle in twodistinct points, is called a secant.

7. A line which has only one point common tothe circle is called a tangent to the circle.

TEXTBOOK’S EXERCISE 10.1

Q.1. How many tangents can a circle have?

Sol. A circle can have infinitely many tangents.

Since there are infinitely many points on a circle and ateach point of it, it has a unique tangent.

Q.2. Fill in the blanks :

(i) A tangent to a circle intersects it in_______ point(s).

(ii) A line intersecting a circle in two pointsis called a _______.

(iii) A circle can have _______ paralleltangents at the most.

(iv) The common point of a tangent to acircle and the circle is called _______.

Sol. (i) A tangent to a circle intersects it in onepoint(s).

(ii) A line intersecting a circle in two points iscalled a secant.

(iii) A circle can have two parallel tangents at themost.

(iv) The common point of a tangent to a circle

and the circle is called point of contact.

Q.3. A tangent PQ at a point P of a circle ofradius 5 cm meets a line through the centre O ata point Q so that OQ = 12 cm. Length PQ is :

(a) 12 cm (b) 13 cm

(c) 8.5 cm (d) 119 cmSol.PQ is the tangent and OP is the radius. OPQ = 90ºIn right angled OPQ,

OQ2 = OP2 + PQ2 (By Pythagoras Theorem)

(12)2 = (5)2 + PQ2

144 = 25 + PQ2

PQ2 = 144 – 25 PQ2 = 119

PQ 119 cm

Hence, the length PQ is 119 cm.

So, the correct choice is (d).

Q.4. Draw a circle and two lines parallel toa given line such that one is a tangent and other,a secant to the circle.

Sol. According to the given information we draw a

circle having O as centre and l is the given line.

10

Question Bank In Mathematics Class X (Term–II)

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Now, m and n be two lines parallel to a given line land that m is tangent as well as parallel to l and n issecant to the circle as well as parallel to l.

TEXTBOOK’S EXERCISE 10.2

In Q.1 to 3 choose the correct option and givejustification.

Q.1. From a point Q, the length of thetangent to a circle is 24 cm and the distance ofQ from the centre is 25 cm. The radius of thecircle is :

(a) 7 cm (b) 12 cm(c) 15 cm (d) 24.5 cmSol. QPO = 90º [Angle between tangent and

radius through the point of contact]In right angled QPO, OQ2 = OP2 + PQ2

[By Pythagoras Theorem]

(25)2 = OP2 + (24)2

625 = OP2 + 576 OP2 = 625 – 576 OP2 = 49

OP 49 7 cm

So, radius of the circle is 7 cm.Hence, the correct option is (a).

Q.2. In figure, if TP and TQ are the twotangents to a circle with centre O so that POQ= 110º, then PTQ is equal to :

(a) 60º (b) 70º(c) 80º (d) 90°

Sol. POQ = 110º (Given)OPT = 90º and OQT = 90º

[Angle between tangent and radius throughthe point of contact is 90º]

In quadrilateral OPTQ,POQ + OPT + OQT + PTQ = 360º

[ The sum of all the angles of a quadrilateralis 360º]

110º + 90º + 90º + PTQ = 360º 290º + PTQ = 360º PTQ = 360º – 290º = 70ºHence, the correct choice is (b).

Q.3. If tangents PA and PB from a point P toa circle with centre O are inclined to each other

10.2 NUMBER OF TANGENTS FROM APOINT ON A CIRCLE

1. There is one and only one tangent at apoint of the circle.

2. The tangent at any point of a circle isperpendicular to the radius through thepoint of contact.

3. No tangent can be drawn from a point

inside the circle.

4. The lengths of tangents drawn from anexternal point to a circle are equal.

5. The perpendicular at the point of contact tothe tangent to a circle passes through thecentre of the circle.

6. Tangents drawn at the end points of adiameter of a circle are parallel.

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at angle of 80º, then POA is equal to :(a) 50º (b) 60º (c) 70º (d) 80°Sol. OAP = 90º [Angle between tangent and

radius through the point of contact]

1OPA BPA

2 [The centre lies on the bisector

of the angle between the two tangents]1

(80º ) 40º2

In OPA, OAP + OPA + POA = 180º[Sum of the three angles of a triangle is 180º]

90º + 40º + POA = 180º 130º + POA = 180ºPOA = 180º – 130ºPOA = 50ºHence, the correct choice is (a).

Q.4. Prove that the tangents drawn at theends of a diameter of a circle are parallel.

[2011 (T-II)]Sol. PQ and RS are tangents to the circle C(O, r) at

points A and B respectively, AB being diameter.Since tangent is perpendicular to the radius at the

point of contact.

AB PQ and AB RSPAB = 90º and ABS = 90ºPAB = ABS PQ || RS [PAB and ABS are

alternate angles] Proved.

Q.5. Prove that the perpendicular at thepoint of contact to the tangent to a circle passesthrough the centre.

Sol. We know that the tangent at any point of acircle is perpendicular to the radius through the point ofcontact and the radius essentially passes through thecentre of the circle, therefore the perpendicular at thepoint of contact to the tangent to a circle passes throughthe centre.

Q.6. The length of a tangent from a point Aat distance 5 cm from the centre of the circle is4 cm. Find the radius of the circle. [Imp.]

Sol. We know that the tangent at any point of a circleis perpendicular to the radius through the point ofcontact.

OPA = 90ºIn right angled OPA, OA2 = OP2 + AP2

[By Pythagoras Theorem]

(5)2 = OP2 + (4)2 25 = OP2 + 16

OP2 = 25 – 16 = 9 OP 9 3 cm

Hence, the radius of the circle is 3 cm.

Q.7. Two concentric circles are of radii 5 cmand 3 cm. Find the length of the chord of thelarger circle which touches the smaller circle.

[Imp.]Sol. Let O be the common centre of the two

concentric circles.Let AB be a chord of the larger circle which

touches the smaller circle at P.Join OP and OA.Then, OPA = 90º[ The tangent at any point of a circle is

perpendicular to the radius through the point of contact]

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In APO, OA2 = OP2 + AP2

[By Pythagoras Theorem] (5)2 = (3)2 + AP2

AP2 = 25 – 9 AP2 = 16

AP 16 4 cm AP = BP = 4 cm

AB = AP + BP = AP + AP = 2 AP = 2(4) = 8 cmHence, the required length is 8 cm.

Q.8. A quadrilateral ABCD is drawn tocircumscribe a circle (see figure). Prove that

AB + CD = AD + BC [2008]Sol. We know that the tangents from an external

point to a circle are equal. AP = AS ....(i)

BP = BQ ...(ii)CR = CQ ...(iii)DR = DS ...(iv)

Adding equations (i), (ii), (iii) and (iv), we have(AP + BP) + (CR + DR)

= (AS + BQ + CQ + DS)

AB + CD = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC. Proved.

Q.9. In figure, XY and XY are two paralleltangents to a circle with centre O and anothertangent AB with point of contact C intersectingXY at A and XY at B. Prove that AOB = 90º.

[HOTS]

Sol. From figure, XY and XY are two paralleltangents to a circle with centre O and another tangent

AB with point of contact C intersecting XY at A andXY at B.

To Prove : AOB = 90ºProof : OPA = 90º ...(i)[The tangent at any point of a circle is

perpendicular to the radius through the point of contact]OCA = 90º ...(ii)

In right triangles OPA and OCA, we haveOA = OA (Common)AP = AC [Tangent segments from an

external point to a circle are equal] OPA OCA [RHS congruence criterion] OAP = OAC [CPCT]

1

OAC PAB2

...(iii)

Similarly, OBQ = OBC

1

OBC QBA2

...(iv)

XY || XYand a transversal AB intersects them. PAB + QBA = 180º[ Sum of the consecutive angles on the same side

of the transversal is 180º]

1 1 1

PAB + QBA (180º )2 2 2

∠ ∠ =

OAC + OBC = 90º ...(v)[From equations (iii) and (iv)]

In AOB,OAC + OBC + AOB = 180º ...(vi)Putting the value of OAC + OBC = 90º in

equation (vi), we have 90º + AOB = 180º AOB = 90º. Proved.

Q.10. Prove that the angle between the twotangents drawn from an external point to a circleis supplementary to the angle subtended by theline-segment joining the points of contact at thecentre. [2011 (T-II)]

Sol. OAP = 90º ...(i)[Angle between tangent and radius through the

point of contact is 90º]

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OBP = 90° ...(ii)

OAPB is a quadrilateral.

APB + AOB + OAP + OBP = 360º

Putting the value of OAP and OBP, we have

APB + AOB + 90º + 90º = 360º

APB + AOB = 180º

APB and AOB are supplementary.

Q.11.Prove that the parallelogram circums-cribing a circle is a rhombus.

[2009, 2011 (T-II)]Sol. Given : ABCD is a parallelogram circums-

cribing circle.To Prove : ABCD is a rhombus.Proof : Since the tangent segments from an

external point to a circle are equal. AP = AS ...(i)

BP = BQ ...(ii)CR = CQ ...(iii)

and DR = DS ...(iv)

Adding equations (i), (ii), (iii) and (iv), we get

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AB + AB = AD + AD

2 AB = 2AD

AB = AD

But AB = CD and AD = BC

[Opposite sides of parallelogram]

AB = BC = CD = AD

Hence, parallelogram ABCD is a rhombus.

Q.12. A triangle ABC is drawn tocircumscribe a circle of radius 4 cm such thatthe segments BD and DC into which BC is dividedby the point of contact D are of length 8 cm and6 cm respectively. (see figure). Find the sides ABand AC. [HOTS]

Sol. Let the sides BC, CA and AB of the ABCtouch the circle at D, F and E respectively.

Join OE and OF. Also join OA, OB and OC.

Now, BD = BE = 8 cm and CF = CD = 6 cm

[Tangent segments from an external point to acircle are equal]

Also, let AE = AF = x cm. Then,a = 14, b = x + 6, c = x + 8

Semi-perimeter of ABC2

a b c

14 ( 6) ( 8)( 14) cm

2

x xx

Area of ABC ( 14)( 14 14)x x

( 14 8)( 14 6) x x x x

( 14)( )(6)(8) x x

Now, area of ABC= Area of OBC + Area of OCA

+ Area of OAB

( 14)( )(6)(8)x x

(6 8)4 ( 6)4 ( 8)4

2 2 2

x x

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( 14)( )(6)(8) 28 2 12 2 16x x x x

( 14)( )(6)(8) 4 56x x x

( 14)( )(6)(8) 4( 14)x x x

Squaring both sides, we have(x + 14)(x)(6)(8) = 16(x + 14)2

3x = x + 14 3x – x = 14

2x = 14 14

72

x

AB = x + 8 = 7 + 8 = 15 cmand AC = x + 6 = 7 + 6 = 13 cm

Q.13. Prove that opposite sides of aquadrilateral circumscribing a circle subtendsupplementary angles at the centre of the circle.

[2011 (T-II)]Sol. Given : ABCD is a quadrilateral circums-

cribing a circle whose centre is O.To Prove : (i) AOB + COD = 180º

(ii) BOC + AOD = 180ºConstruction : Join OP, OQ, OR and OS.Proof : Since tangent segments from an external

point to a circle are equal, thereforeAP = AS, BP = BQ, CQ = CR, DR = DSIn OPB and OBQ,

OP = OQ [Radii of the same circle]

OB = OB [Common]BP = BQ [Proved above]

OBP OBQ [SSS congruence criterion] 1 = 2 [CPCT]

Similarly, 3 = 4, 5 = 6 and 7 = 8So, 1 + 2 + 3 + 4 + 5 + 6 + 7 +

8 = 360º 1 + 1 + 4 + 4 + 5 + 5 + 8 +

8 = 360º 2(1 + 4 + 5 + 8) = 360º 1 + 4 + 5 + 8 = 180º (1 + 8) + (4 + 5) = 180º AOB + COD = 180ºSimilarly, we can prove thatBOC + AOD = 180º. Proved.

OTHER IMPORTANT QUESTIONS

Q.1. The common point of the tangent andthe circle is called the :

(a) point of contact(b) point of intersection(c) point of concurrence(d) none of the aboveSol. (a) The common point of the tangent and the

circle is called the point of contact.

Q.2. How many tangents of a circle passingthrough a point lying inside the circle?

(a) zero (b) one(c) two (d) infiniteSol. (a) There is no tangent to a circle passing

through a point lying inside the circle.

Q.3. How many tangents can we draw to acircle from a point lying outside the circle?

(a) 0 (b) 1 (c) 2 (d) infinite

Sol. (a) There are exactly 2 tangents to a circle

through a point lying outside the circle.

Q.4. The tangents drawn at the ends of adiameter of a circle are :

(a) parallel (b) perpendicular(c) inclined at an angle of 60º(d) none of the aboveSol. (a) The tangents drawn at the ends of a

diameter of a circle are parallel.

Q.5. In the given figure, PT is a tangent tothe circle with centre O. If OT = 6 cm and

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OP = 10 cm, then the length of tangent PT is :

(a) 8 cm (b) 12 cm(c) 10 cm (d) 16 cmSol. (a) By Pythagoras Theorem in OPT,

PT2 = OP2 – OT2

PT2 = (10)2 – (6)2 = 64 PT = 8 cm

Q.6. In a circle of radius 7 cm, tangent PTis drawn from a point P such that PT = 24 cm.If O is the centre of the circle, then length of OPis :

(a) 30 cm (b) 28 cm(c) 25 cm (d) 31 cmSol. (c) OP2 = PT2 + OT2 = (24)2 + (7)2

= 576 + 49 = 625

OP 625 25 cm

Q.7. A point P is 26 cm away from the centreof a circle and the length of tangent drawn fromP to the circle is 24 cm. The radius of the circleis :

(a) 8 cm (b) 10 cm(c) 12 cm (d) 14 cmSol. (b) Let O be the centre of the circle and PT be

the tangent. Then, radius of the circle is OT.OT2 = OP2 – PT2 = (26)2 – (24)2

= 676 – 576 = 100

OT2 = 100 OT 100 10 cm

Q.8. In the given figure, O is the centre oftwo concentric circles of radii 3 cm and 5 cm.AB is a chord of outer circle which touches theinner circle. The length of chord AB is :

(a) 4 cm (b) 8 cm

(c) 34 cm (d) 7 cm

Sol. (b) Draw OP AB. Join OA.

Then, OP = 3 cm and OA = 5 cm.OA2 = OP2 + AP2

AP2 = OA2 – OP2 = 52 – 32

AP2 = 16 AP = 16 4 cm

AB = 2 × AP = (2 × 4) cm = 8 cm

Q.9. In the figure, ABC is circumscribedtouching the circle at P, Q, R. If AP = 4 cm, BP= 6 cm, AC = 12 cm and BC = x cm. Then, x = ?

(a) 10 cm (b) 6 cm

(c) 14 cm (d) 18 cm

Sol. (c) Since, the lengths of tangents drawn froman external point to a circle are equal, thenAR = AP = 4 cm

CR = (AC – AR) = (12 – 4) cm = 8 cm = CQAnd, BQ = BP = 6 cm BC = BQ + CQ = (6 + 8) cm = 14 cm.

Q.10. In the given figure, quad. ABCD iscircumscribed, touching the circle at P, Q, R andS. If AP = 5 cm, BC = 7 cm and CS = 3 cm, thenlength AB = ?

(a) 9 cm (b) 10 cm(c) 12 cm (d) 8 cmSol. (a) Since the lengths of tangents drawn from

an external point to a circle are equal, then

AQ = AP = 5 cm

CR = CS = 3 cm, BR = (BC – CR)

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= (7 – 3) cm = 4 cm

BQ = BR = 4 cm

AB = AQ + BQ = (5 + 4) cm = 9 cm.

Q.11. In the given figure, ABCD iscircumscribed touching the circle at P, Q, R andS. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR= 4 cm, then perimeter of quad. ABCD is :

(a) 18 cm (b) 27 cm(c) 36 cm (d) 22 cmSol. (c) Since the length of tangents from an

external point to a circle are equal, thenAP = AS = 6 cm, BP = BQ = 5 cm,

CR = CQ = 3 cm, DS = DR = 4 cm Perimeter of quad. ABCD= (AP + BP) + (BQ + CQ) + (CR + DR)

+ (AS + DS) = 36 cm.

Q.12. In the given figure, quad. ABCD iscircumscribed, touching the circle at P, Q, R andS such that DAB = 90º. If CS = 27 cm and CB= 38 cm and the radius of the circle is 10 cm,then AB = ?

(a) 17 cm (b) 28 cm(c) 19 cm (d) 21 cmSol. (d) Since the lengths of tangents drawn from

an external point to the circle are equal, thenCR = CS = 27 cm.So, BR = (BC – CR) = (38 – 27) cm = 11 cm BQ = BR = 11 cm.Join OQ. Then, PAQO is a rectangle. AQ = PQ = 10 cm.Hence, AB = (AQ + BQ) = (10 + 11) cm = 21 cm.

Q.13. In the given figure, ABC is right-angled at B such that BC = 6 cm and AB = 8 cm.A circle with centre O has been inscribed insidethe triangle. OP AB, OQ BC andOR AC. If OP = OQ = OR = x cm, then x = ?

[2011 (T-II)]

(a) 2 cm (b) 3 cm(c) 2.5 cm (d) 4 cmSol. (a) AC2 = (AB2 + BC2) = 82 + 62

= 64 + 36 = 100

AC = 100 = 10 cmNow, CR = CQ = BC – BQ = (6 – x) cmand AR = AP = (AB – BP) = (8 – x) cm

AC = CR + AR = (6 – x) + (8 – x)= (14 – 2x)

14 – 2x = 10 x = 2 cmHence, radius of the circle = 2 cm

Q.14. Quadrilateral ABCD circumscribes acircle as shown in figure. The side of thequadrilateral which is equal to AP + BR is :

[2011 (T-II)](a) AD (b) AC(c) AB (d) BC

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Sol. (c) Since the lengths of tangents drawn from

an external point to the circle are equal, thenAP = AQ and BR = QB.

AP + BR = AQ + QB = AB [AQ + QB = AB]

Q.15. In figure, if OC = 9 cm and OB = 15cm, then BC + BD is equal to : [2011 (T-II)]

(a) 18 cm (b) 12 cm(c) 24 cm (d) 36 cmSol. (c) In OCB, by Pythagoras theorem

CB = 2 2(15) (9) 225 81 144− = − =CB = 12 cmBD = BC = 12 cm [ length of tangents

drawn from an external point to the circle are equal]Now, BC + BD = 12 cm + 12 cm

= 24 cm

Q.16. A tangent PA is drawn from an

external point P to a circle of radius 3 2 cm

such that the distance of the point P from O is6 cm as shown in figure. The value of APO is :

[2011 (T-II)]

(a) 30° (b) 45°(c) 60° (d) 75°Sol. (b) Let APO = .

In right OAP, we have

sin =OA

OP

3 2 1sin

6 2⇒ θ = =

45⇒ θ = °

Q.17. The figure, shows two concentriccircles with centre O. AB and APQ are tangentsto the inner circle from point A lying on the outer

circle. If AB = 7.5 cm, then AQ is equal to :[2011 (T-II)]

(a) 18 cm (b) 15 cm(c) 12 cm (d) 10 cmSol. (b) AP = AB = 7.5 cm [since the length of

tangents from an externalpoint to a circle are equal]

Now, AQ = 2AP = 2 × 7.5 cm = 15 cm[since perpendicular from the centre bisects the chord]

Q.18. In figure, APB is a tangent to a circlewith centre O, at point P. If QPB = 50°, thenthe measure of POQ is : [2011 (T-II)]

(a) 120° (b) 100° (c) 140° (d) 150°Sol. (b) OPB = 90° [ the tangent at any point

of a circle is perpendicular to the radius through thepoint of contact]

OPQ = 90° – QPB = 90° – 50° = 40°Now, since OP = OQ (radii)OQP = OPQ = 40°40° + 40° + POQ = 180°POQ = 180° – 80° = 100°

Q.19. In the given figure, the length of PRis : [2011 (T-II)]

(a) 20 cm (b) 26 cm(c) 24 cm (d) 28 cm

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Sol. (b) In right OSR, OR2 = 52 + 122 = 169

OR = 169 OR = 13 cm

In right PQO,

PO2 = 42 + 32 = 25 PO = 25

PO = 5 cmNow, OO = (3 + 5) cm = 8 cmPR = PO+ OO + OR = 5 cm + 8 cm + 13 cm

= 26 cm

Q.20. CP and CQ are tangents to a circlewith centre O. ARB is another tangent touchingthe circle at R. If CP = 11 cm, BC = 7 cm, thenthe length BR is : [2011 (T-II)]

(a) 11 cm (b) 7 cm(c) 3 cm (d) 4 cmSol. (d) CP = CQ [ length of tangents from

an exterior point to a circle are equal]11 cm = QB + 7 cmQB = 11 cm – 7 cm = 4 cm BR = QB = 4 cm [ length of tangents from

an exterior point to a circle are equal]

Q.21. In the given figure, O is the centre ofthe circle. If PA and PB are tangents from anexternal point P to the circle, then AQB isequal to : [2011 (T-II)]

(a) 100° (b) 80°(c) 70° (d) 50°

Sol. (d) PAO = PBO = 90°

[Angle between tangent andradius through the point of contact]

AOB = 360° – (80° + 90° + 90°)= 360° – 260° = 100°

Now, AQB =1

2AOB

[ The angle subtended by an arc at the centre isdouble the angle subtended by it at any point on theremaining part of the circle]

1100 50

2= × ° = °

Q.22. In the given figure, if AOB = 125°,then COD is equal to : [2011 (T-II)]

(a) 62.5° (b) 45°(c) 125° (d) 55°

Sol. (d) If a circle touches the sides of a

quadrilateral, then the angles subtended at the centre bya pair of opposite sides are supplementary.

AOB + COD = 180°COD = 180° – 125° = 55°

Q.23. In the given figure, AT is the tangent to thecircle with centre O such that OT = 4 cm andOTA = 30°. Then AT is equal to : [2011 (T-II)]

(a) 4 cm (b) 2 cm

(c) 2 3 cm (d) 8 cm

Sol. (c) Let AT = x cm

cos 30° =3

4 2 4⇒ =x x

x = 2 3 cm

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Q.24. In the given figure, CP and CQ aretangents from an external point C to a circle withcentre O. AB is another tangent which touchesthe circle at R. If CP = 11 cm and BR = 4 cm,find the length of BC. [2010]

Sol. We have, CP = 11 cm, BR = 4 cm

Since, the tangent from an external point are equal,we have CQ = CP = 11 cm and BQ = BR = 4 cm

BC = CQ – BQ = (11 – 4) cm = 7 cm

Q.25. In the given figure, there are twoconcentric circles with centre O and of radii 5cm and 3 cm. From an external point P, tangentsPA and PB are drawn to these circles. If AP =12 cm, find the length of BP. [2010]

Sol. Join OA, OB and OP.

Then OA = 5 cm, OB = 3 cm and PA = 12 cm

In OPA, OP2 = OA2 + AP2

= 52 + 122

= 25 + 144 = 169

OP = 169 = 13 cm

Now, in OBP, OP2 = OB2 + BP2

BP2 = OP2 – OB2 = (13)2 – (3)2

= 160

= BP = 160 cm = 4 10 cm

Q.26. In the given figure, if ATO = 40º,find AOB. [2008]

Sol. In OAT and OBT,

4 = 1 = 90º [Tangent is to theradius through the point of contact]

OT = OT [Common]OA = OB [Radius]

OAT OBT [RHS criterion] 3 = 2 ....(i) [CPCT]

In OAT, 3 + 4 + 5 = 180º[Angle sum property of a ]

3+ 90º + 40º = 180º3 + 130º = 180º 3 = 50º AOB = 2 + 3 = 3 + 3 [From (i)]

= 50º + 50º = 100º.

Q.27. From a point P, the length of thetangent to a circle is 15 cm and distance of Pfrom the centre of the circle is 17 cm. Then whatis the radius of the circle? [2008C]

Sol. OAP = 90º

[Radius is to the tangent through the point ofcontact]

In right OAP, OA2 + AP2 = OP2

[Pythagoras Theorem]

r2 + (15)2 = (17)2 r2 + 225 = 289 r2 = 289 – 225 = 64 r = 8 cm.

Q.28. The two tangents from an externalpoint P to a circle with centre O are PA and PB.

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If APB = 70º, what is the value of AOB?[2008C]

Sol. 1 = 2 = 90º

[Tangent is to the radius through the point ofcontact]

In quad. OAPB,AOB + 1 + APB + 2 = 360º

[Angle sum property of a quad.]AOB + 90º + 70º + 90º = 360ºAOB + 250º = 360ºAOB = 360º – 250º = 110º.

Q.29. In the figure, ABC is circumscribinga circle. Find the length of BC. [2009]

Sol. We know that the lengths of tangents drawn

from an external point to a circle are equal.So, AR = AQ = 4 cm ...(i)

BR = BP = 3 cm ...(ii)CP = CQ = (AC – AQ) = (11 – 4) cm

= 7 cmSo, BC = BP + CP = (3 + 7) cm = 10 cm.

Q.30. In the figure, CP and CQ are tangents

to a circle with centre O. ARB is another tangent

touching the circle at R. If CP = 11 cm and

BC = 7 cm, then find the length of BR. [2009]

Sol. We know that the lengths of tangents drawn

from an external point to a circle are equal.

So, CP = CQ ... (i)

AP = AR ... (ii)

BQ = BR ... (iii)

Now, CP = CQ

11= BC + BQ

11= 7 + BR [From (iii)]

BR= 11 – 7 = 4 cm.

Q.31. Two tangents PA and PB are drawn to

a circle with centre O from an external point P.

Prove that APB = 2OAB.

[2009, 2011 (T-II)]

Sol. OAP = 90° [Radius through the point

of contact is perpendicular to the tangent]

PAB = 90° – OAB

In APB, PA = PB [Tangents from an

external point to a circle are equal]

PBA = PAB = 90° – OAB ...(i)

Also, APB = 180° – (PBA + PAB)

= 180° – 2 (90° – OAB)

APB = 180° – 180° + 2OAB

APB = 2OAB Proved.

Q.32. ABC is an isosceles triangle, in which

AB = AC, circumscribed about a circle. Show

that BC is bisected at the point of contact.

[2008, 2011 (T-II)]

Sol. Given : An isosceles triangle ABC, in which

AB = AC, circumscribed about a circle.

To prove : BF = FC

Prove : We know that the tangents drawn from an

external point to a circle are equal.

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So, AD = AE …(i)

CE = CF …(ii)

BD = BF …(iii)

Also, AB = AC [Given]

AD+ BD = AE + CE

DB = CE …(iv) [From (i)]

From (ii), (iii) and (iv), we have BF = FC

Hence, BC is bisected at the point of contact.

Proved.

Q.33. In the figure a circle is inscribed in aquadrilateral ABCD in which B = 90°. If AD =23 cm, AB = 29 cm and DS = 5 cm, find theradius (r) of the circle. [2008]

Sol. We know that the tangents drawn from an

external point to a circle are equal. AR = AQ …(i)

DR = DS …(ii)BQ = BP …(iii)AD = AR + DR

AD = AR + DS [From (ii)] 23 = AR + 5 AR = 23 – 5 = 18 cm

AB = AQ + BQ AB = AR + BQ [From (i)] 29 = 18 + BQ BQ = 29 – 18 = 11 cm …(iv)In quadrilateral OPBQ,

B = 90°OQB = OPB = 90°

[Tangent is perpendicular to the radius at the pointof contact]

BQ = BP [From (iii)]OQ = OP = r [Given]

OPBQ is a square. OP = OQ = PB = BQ r = 11 cm [From (iv)]

Q.34. In the figure, OP is equal to diameterof the circle. Prove that ABP is an equilateraltriangle. [2008, 2011 (T-II)]

Sol. Suppose OP meets the circle at Q. Join QA

and AB.

We have, OP = diameter

OQ + PQ = diameter

PQ = diameter – radius.

PQ = radius

So, OQ = PQ = radius

Thus, OP is the hypotenuse of right triangle OAPand Q is the mid-point of OP.

OA = AQ = OQ

[ Mid-point of hypotenuse of a right triangle isequidistant from the vertices]

OAQ is an equilateral triangle.AOQ = 60°So, APO = 30° APB = 2APO = 60°Also, PA = PB PAB = PBABut, APB = 60°Therefore, PAB = PBA = 60°Hence, APB is equilateral. Proved.

Q.35. In the figure, ∠ADC = 90°, BC =38 cm, CD = 28 cm and BP = 25 cm. Find theradius of the circle. [2006, 2011 (T-II)]

P

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Sol. Since tangent to a circle is perpendicular to

the radius through the point of contact.∴ ∠OSD = ∠ORD = 90°, OR = OS⇒ DROS is a square.Als,o BP = BQ

[Tangents from an external point are equal]

⇒ BQ = 25 cm [BP = 25 cm]

⇒ BC – CQ = 25

⇒ 38 – CQ = 25 [BC = 38 cm]

⇒ CQ = 38 – 25 = 13 cm

⇒ CR = CQ = 13 [CQ = 13 cm]

⇒ CD – DR = 13 [CR = CD – DR]

⇒ 28 – DR = 13 [CD = 28 cm]

⇒ DR = 28 – 13 = 15 cm

Since DROS is a square, so OR = DR = 15 cm.

Hence, radius of the circle = 15 cm.

Q.36. PQR is a right-angled triangle withQR = 12 cm and PQ = 5 cm. A circle with centreO is inscribed in ∆PQR. Find the radius of thecircle.

Sol.

Since OA ⊥ PQ, OB ⊥ QR and OA = OB

⇒ AOBQ is a square.

In right-angled triangle PQR,

PR2 = PQ2 + QR2

[Pythagoras theorem]

⇒ PR2 = 52 + 122

⇒ PR2 = 25 + 144

⇒ PR2 = 169

⇒ PR = 13 cm

PA = PC [Tangents drawnfrom an external point are equal]

⇒ PQ – AQ = PR – RC

⇒ 5 – r = 13 – RC

⇒ RC = 13 – 5 + r

⇒ RC = 8 + r

RB = RC [Tangents drawnfrom an external point are equal]

⇒ QR – BQ = RC ⇒ 12 – r = 8 + r

⇒ 2r = 4 ⇒ r = 2 cm

Hence, radius of the circle is 2 cm.

Q.37. A circle touches the side BC of ∆ABC,at P and touch AB and AC produced at Q and R

respectively. Prove that AQ =1

2Perimeter of

∆ABC. [2006, 2011 (T-II)]Sol.Since tangents from an exterior point to a

circle are equal in length.

∴ BP = BQ ...(i)

CP = CR ...(ii)

and AQ = AR ...(iii)

Now, perimeter of ∆ABC

= AB + BC + AC

= AB + (BP + PC) + AC

= (AB + BQ) + (AC + CR)[From (i) and (ii)]

= AQ + AR = 2AQ [ AQ = AR]

So, AQ =1

2(Perimeter of ∆ABC).

Q.38. In the figure, O is the centre of thecircle, PA and PB are tangent segments. Showthat the quadrilateral AOBP is cyclic.

[2011 (T-II)]Sol. OA ⊥ AP and OB ⊥ BP

[Tangent to a circle is perpendicular to the radius atthe point of contact.]

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⇒ ∠OAP = 90° and ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180° ...(i)

⇒ OAPB is cyclic because in a quadrilateral if apair of opposite angles are supplementary, then thequadrilateral is cyclic. Proved.

Q.39. In the figure, circles C(O, r) and

C(O′,2

r) touch internally at a point A and AB is

a chord of the circle C(O, r), intersecting

C(O′,2

r) at C. Prove that AC = CB. [HOTS]

Sol. Join OC, OA.

∠OCA is the angle in a semicircle.∴ ∠OCA = 90° ... (i)⇒ OC ⊥ ABIn circle C(O, r), AB is chord and OC ⊥ AB

[From (i)]∴ AC = CB

[Perpendicular from centre of circle to a chordbisects the chord] Proved.

Q.40. If two tangents are drawn to a circlefrom an external point, then (i) they subtendsequal angles at the centre, (ii) they are equallyinclined to the segment joining the centre to thatpoint. Prove. [HOTS]

Sol.

Given : PT and PT′ are two tangents drawn to acircle C (O, r) from an external point P.

To prove : (i) ∠POT = ∠POT′ and

(ii) ∠OPT = ∠OPT′Proof : In ∆POT and ∆POT′

PT = PT′[Tangents drawn from an external point]

OT = OT′ [Radii of the same circle]

∠PTO = ∠PT′O [Each = 90°]

⇒ ∆POT ≅ ∆POT′ [SAS congruency]

⇒ ∠POT = ∠POT′and ∠OPT = ∠OPT′ [CPCT]

Q.41. In the figure below, the incircle of∆ABC touches the sides BC, CA and AB at D, Eand F respectively. Show that :

AF + BD + CE = AE + BF + CD

=1

2(Perimeter of ∆ABC)

[Imp.]

Sol. We know that the tangents from an exterior

point to a circle are equal in length.

∴ AF = AE ... (i)

BD = BF ... (ii)

CE = CD ... (iii)

Adding (i), (ii) and (iii), we get

AF + BD + CE = AE + BF + CD

Perimeter of ∆ABC = AB + BC + AC

= (AF + FB) + (BD + CD) + (AE + EC)

= (AF + AE) + (BF + BD) + (CD + CE)

= (2AF + 2BD + 2CE)

= 2(AF + BD + CE) [From (i), (ii) and (iii)]

AF + BD + CE =1

2(Perimeter of ∆ABC)

Hence, AF + BD + CE = AE + BF + CD

=1

2(Perimeter of ∆ABC) Proved.

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Q.42. In the figure, if AB = AC, prove thatBE = EC. [2011 (T-II)]

Sol. We know that the tangents from an exterior

point to a circle are equal in length.

⇒ AD = AF ...(i)

BD = BE ...(ii)

CE = CF ...(iii)

Now, AB = AC

⇒ AB – AD = AC – AD

[Subtracting AD from both sides]

⇒ AB – AD = AC – AF [Using (i)]⇒ BD = CF⇒ BE = CE Proved.

[Using (ii), (iii)]

Q.43. PQ is a chord of length 8 cm of acircle of radius 5 cm. The tangents at P and Qintersect at a point T. Find the length TP.

[HOTS]Sol. Let TR = y.

Since OT is perpendicular bisector of PQ.

PR = QR = 4 cm

In right triangle ORP, we have

OP2 = OR2 + PR2

OR2 = OP2 – PR2 = 52 – 42 = 9

OR = 3 cm

In right triangles PRT and OPT, we have

TP2 = TR2 + PR2 ...(i)

and, OT2 = TP2 + OP2

OT2 = (TR2 + PR2) + OP2 [From (i)]

(y + 3)2 = y2 + 16 + 25

6y = 32

y =16

3

TR =16

3

From (i), we have

TP2 =

2216 256 400

4 163 9 9

TP =20

3cm

Q.44. Let A be one point of intersection oftwo intersecting circles with centres O and Q.The tangents at A to the two circles meet thecircles again at B and C, respectively. Let thepoint P be located so that AOPQ is aparallelogram. Prove that P is the circumcentreof the triangle ABC. [HOTS]

Sol. If P is the circumcentre of ABC, then OP and

PQ are perpendicular bisectors of AB and ACrespectively. i.e., we have to show that OP and PQ areperpendicular bisectors of AB and AC respectively.

We have, OA AC

[ Tangent to a circle is to the radius throughthe point of contact]

PQ AC

[ OAQP is a parallelogram OA || PQ]

PQ is the perpendicular bisector of AC

[ Q is the centre of the circle]

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Similarly,

BA AQ

BA OPAQPO is a parallelogram

OP || AQ

OP is the perpendicular bisector of AB.

Thus, P is the point of intersection of perpendicularbisectors PQ and PO of sides AC and AB respectively.

Hence, P is the circumcentre of ABC. Proved.

Q.45. If AB, AC, PQ are tangents in figurebelow and AB = 5 cm, find the perimeter ofAPQ. [2011 (T-II)]

Sol. We have, PB = PX, QC = QX and AB = AC

Perimeter of APQ= AP + PQ + AQ= AP + (PX + XQ) + AQ= (AP + PX) + (AQ + XQ)= (AP + PB) + (AQ + QC)= AB + AC = (5 + 5) cm = 10 cm

Q.46. In figure below, PQ is tangent at apoint R of the circle with centre O.

If TRQ = 30°, find PRS. [Imp.]

Sol. We have, TRQ = 30°

Since ST is a diameter and angle in a semicircleis a right angle.

SRT = 90°

Now, TRQ + SRT + PRS = 180°

[Straight angle] 30° + 90° + PRS = 180°

PRS = 180° – 120° = 60°.

Q.47. In figure below, a circle touches all thefour sides of a quadrilateral ABCD with AB = 6cm, BC = 7 cm and CD = 4 cm. Find AD.

[2011 (T-II)]

Sol. Since tangents drawn from an exterior point to

the circle are equal.

AP = AS ...(1)

BP = BQ ...(2)

CR = CQ ...(3)

and DR = DS ...(4)

Adding (1), (2), (3) and (4), we get

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

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

6 + 4 = AD + 7 AD = 3 cm.

Q.48. In the figure, two circles touch eachother externally at C. Prove that the commontangent at C bisects the other two commontangents. [2007, 2011 (T-II)]

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Sol. Given : Two circles with centres at A and B

touch each other externally at C. RS and PQ are twocommon tangents to the circles.

To prove : (i) PE = EQ(ii) RF = FS.

Proof : We know that the tangents drawn from anexterior point are equal.

PE = EC …(1)Similarly, EQ = EC …(2)

PE = EQ [From (1) and (2)] EC bisects PQ at E.Similarly, we can prove that F is the bisector of RS.So, EF bisects PQ and RS. Proved.

Q.49. Prove that the lengths of tangentsdrawn from an external point to a circle areequal. Using the above, prove the following :

A quadrilateral ABCD is drawn tocircumscribe a circle (see fig.). Prove thatAB + CD = AD + BC. [2008]

Sol. Part I

Given : PT and PT′ are two tangents drawn from anexternal point P to a circle C (O, r).

To prove : PT = PT

Construction : Join TO, TO and PO.

Proof : In POT and POT

PO = PO (Common)

OT = OT (Radii of the same circle)

PTO = PTO = 90° [OT PT and OT PT' ]

POT POT (RHS congruence)

PT = PT (CPCT) Proved.

Part II : See Q.8 (Textbook’s Exercise 10.2)

Q.50. Prove that the lengths of the tangentsdrawn from an external point to a circle areequal.

Using the above, do the following :

In the figure, TP and TQ are tangents from Tto the circle with centre O and R is any point onthe circle. If AB is a tangent to the circle at R,prove that

TA + AR = TB + BR. [2008, 2011 (T-II)]

Sol. Part I : Same as Q.49.

Part IIFrom part I, we have

TP = TQ …(i)

AP = AR …(ii)

BQ = BR …(iii)

From (i), TA + AP = TB + BQ

TA + AR = TB + BR [From (ii) and (iii)]

Proved.

Q.51. Prove that the tangent at any point ofa circle is perpendicular to the radius throughthe point of contact.

Using the above, do the following :

In the figure, O is the centre of the twoconcentric circles. AB is a chord of the largercircle touching the smaller circle at C. Provethat AC = BC. [2009, 2011 (T-II)]

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1. A point P is 13 cm from the centre of acircle. Radius of the circle is 5 cm. Then,the length of the tangent drawn from P to thecircle is :

(a) 10 cm (b) 11 cm(c) 12 cm (d) 13 cm2. The length of the tangent drawn from a

point, whose distance from the centre of a circleis 20 cm and radius of the circle is 16 cm is :

(a) 12 cm (b) 144 cm(c) 169 cm (d) 25 cm3. A tangent PQ at a point P of a circle of

radius 15 cm meets a line through the centre Oat a point Q so that OQ = 25 cm. Length of PQis :

(a) 5 cm (b) 25 cm(c) 16 cm (d) 20 cm4. In the figure, if AD, AE and BC are

tangents to the circle at D, E and F respectively,then :

(a) AD = AB + BC + CA(b) 2AD = AB + BC + CA

(c)AD

2= AB + BC + CA

(d)AD

4= AB + BC + CA

5. Two circles touch each other externally atC and AB is a common tangent to the circles.ACB is :

(a) 60° (b) 45° (c) 90° (d) 180°

6. PQ is a tangent drawn from a point P toa circle with centre O and QOR is a diameter ofthe circle such that POR = 120°. OPQ is :

(a) 50° (b) 40° (c) 30° (d) 25°

7. In the figure, a circle touches all the foursides of a quadrilateral ABCD, whose three sidesare AB = 6 cm, BC = 7 cm and CD = 4 cm. FindAD. [2011 (T-II)]

PRACTICE EXERCISE 10.1A

Sol. Part I :

Given : A circle (O, r) and a tangent XY at P.To prove : OP ⊥ XYConstruction : Take any point Q on XY. Join OQ

which intersects the circle at R.Proof : OP = OR (Radii of the circle)Now, OQ = OR + RQ OQ > OR

OQ > OP (OP = OR)

OP < OQ

Thus, OP is shorter than any other point on XY.

Hence, OP XY

(Perpendicular distance is the shortest distancefrom a point to a line) Proved.

Part II :

Join OC. Then,OCB = 90° [From part I]

For outer circle, AB is a chord

AC = BC

[Perpendicular from the centre bisects the chord]Proved.

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8. From an external point P, tangents PA andPB are drawn to a circle with centre O. If CD isthe tangent to the circle at a point E andPA = 14 cm, find the perimeter of PCD.

9. In the given figure, O is the centre of twoconcentric circles of radii 4 cm and 6 cmrespectively. PA and PB are tangents to the outerand inner circles respectively. If PA = 10 cm, findthe length of PB. [2011 (T-II)]

10. PA and PB are tangents from an exteriorpoint P to a circle such that PA = 10 cm andAPB = 60°. Find the length of chord AB.

11. In PQR, Q = 90°. A circle with centreO and radius 2 cm has been inscribed in thetriangle. If QR = 6 cm, find the perimeter ofPQR. [Imp.]

12. In the figure, if AB = 8 cm, BC = 7 cmand CA = 5 cm, find BX.

13. In the figure, a circle is inscribed inABC having sides 8 cm, 10 cm and 12 cm. FindAD, BE and CF. [2011 (T-II)]

14. In the figure, two circles with centres Oand O and radii 8 cm and 4 cm touch each otherexternally. Find the length of their commontangent QR. [HOTS]

15. In the figure, OP and OQ are two radii ofa circle and TP and TQ are tangents to the circle.If POQ = 40°, find PTQ. [HOTS]

16. In the figure, OT is a radius and PT is atangent to the circle. If OT = 12 cm and PT = 16cm, find PQ. [Imp.]

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17. In the figure, O is the centre of the circle,AB is a diameter and AD is a tangent. If BOC= 120°, find ADC.

18. In the figure, PA and PB are the tangentsegments to a circle with centre O. Show that A,O, B, P are concyclic.

19. Prove that in two concentric circles, thechord of the larger circle which touches thesmaller circle is bisected at the point of contact.

20. Prove that the line segment joining thepoint of contact of two parallel tangents to acircle is a diameter of the circle. [2011 (T-II)]

21. Prove that there is one and only onetangent at any point on the circumference of acircle.

22. From a point P, two tangents PA and PBare drawn to a circle C(O, r). If OP = 2r, showthat APB is equilateral. [Imp.]

23. In the given figure PO QO. Thetangents to the circle with centre O at P and Qintersect at a point T. Prove that PQ and OT areright bisectors of each other. [2011 (T-II)]

24. If from an external point B of a circlewith centre O, two tangents BC and BD aredrawn such that DBC = 120°, prove thatBO = 2BC. [2011 (T-II)]

B. FORMATIVE ASSESSMENT

Activity

Objective : To show the following with the help of an activity.

(a) Lengths to tangents drawn from an external point are equal.

(b) Tangents are equally inclined to the segment joining the centre to that point.

Materials Required : White sheet of paper (or rice paper), Geometry box, Sketch pens, A pairof scissors.

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Figure 2

Figure 1

Figure 3

Preparation for the Activity

1. Draw a circle of any radius with centre 'O' on a rice paper.

2. Take a point, say P, outside of the circle.

3. From point P, draw a line segment touching the circle at A(the point of contact), which is the required tangent.[See figure 1]

4. Following step 3, draw one more tangent PB, B being thepoint of contact. [See figure 1]

5. Join OA, OP and OB. [See figure 1]

(a) Lengths of tangents drawn from an external point are equal.

Presentation : Fold the paper along OP.

Observation :

You will observe that point A coincides with point B andline segment PA (the tangent from point P) concides with linesegment PB (another tangent from the same point on the circle).[See figure 2]

Thus, length of PA = length of PB.

Hence, lengths of tangents drawn from an external point are equal.

The above property can be verified a number of times by taking external point at differentlocations.

Result : Lengths of tangents drawn from an external point are equal.

(b) Tangents are equally inclined to the line segment joining the centre with the externalpoint.

1. Cut out the two triangles, OPA and OPB, so formed figure 3.

2. Colour the two triangles with different colours.

3. Put one triangle on the other.

Observation :

You will observe that two triangles are congruent to each other(i.e., one triangle exactly superimposes the other) with thefollowing (angle) correspondence.

OPA = OPB, (Proves the required result)

PAO = PBO,

AOP = BOP (See figure 4)

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Figure 4

Hence, tangents drawn from an external point, equallyinclined to the line segment joining the centre with that point.

Result : Tangents drawn equally inclined to the line segmentjoining the centre with the external point.

Note : The tangent at any point of a circle is perpendicular tothe radius through the point of contact OAP = OBP = 90º.

ANSWERS

A. SUMMATIVE ASSESSMENT

Practice Exercise 10.1A

1. (c) 2. (a) 3. (d) 4. (b) 5. (c) 6. (c)7. 3 cm 8. 28 cm 9. 10.9 cm 10. 10 cm 11. 24 cm 12. 5 cm

13. 7 cm, 5 cm, 3 cm 14. 8 2 cm 15. 140º 16. 8 cm 17. 60º