## Circle

**NCERT TEXTBOOK QUESTIONS SOLVED**

**EXERCISE 10.1**

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

**Sol.**A circle can have an infinite number of tangents.

**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 points is called a……….. .

(iii) A circle can have………..parallel tangents at the most.

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

**Sol.**(i) exactly one (ii) secant (iii) two (iv) point of contact.

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

(A) 12 cm

(B) 13 cm

(C) 8.5

cm (D)

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

**Sol.**We have the required figure.

Here, l is the given line and a circle with centre 0 is drawn.

The line PT is drawn which is parallel to l and tangent to the circle.

Also, AB is drawn parallel to line l and is a secant to the circle.

**EXERCISE 10.2**

**Q.1.**Choose the correct option:

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

**Sol.**QT is a tangent to the circle at T and OT is

radius

Also, OQ = 25 cm and QT = 24 cm

? Using Pythagoras theorem, we get

OQ

^{2}= QT^{2}+ OT^{2} ? OT

^{2}= OQ^{2}– QT^{2} = 25

^{2}– 24^{2}= (25 – 24) (25 + 24) = 1 × 49 = 49 = 7

^{2} ? OT = 7

Thus, the required radius is 7 cm.

? The correct option is (A).

**Q.2.**Choose the correct option:

In figure, if TP and TQ are the two tangents to a circle with centre O so that LPOQ = 110°, then LPTQ is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

**Sol.**?TQ and TP are tangents to a circle with centre O.

such that ?POQ = 110°

? OP ? PT and OQ ? QT

? ?OPT = 90° and ?OQT = 90°

Now, in the quadrilateral TPOQ, we get

? ?PTQ + 90° + 110° + 90° = 360°

? ?PTQ + 290° = 360°

? ?PTQ = 360° – 290° = 70°

Thus, the correct option is (B).

**Q.3.**Choose the correct option:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ?POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

**Sol.**Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.

? OA ? AP and OB ? BP

? ?OAP = ?OBP = 90°

Now, in quadrilateral PAOB, we have:

?APB + ?PAO + ?AOB + ?PBO = 360°

? 80° + 90° + ?AOB + 90° = 360°

? 260° + ?AOB = 360°

? ?AOB = 360° – 260°

? ?AOB = 100°

In rt ?OAP and rt ?OBP, we have

OP = OP

?OAF = ?OBP

OA = OB

? ? OAP ? ? OBP

? Their corresponding parts are equal

? ?POA = ?POB

Thus, the option (A) is correct.

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

**Sol.**In the figure, we have:

PQ is diameter of the given circle and O is its centre.

Let tangents AB and CD be drawn at the end points of the diameter PQ.

Since the tangent at a point to a circle is perpendicular to the radius through the point.

PQ ? AB ? ?APQ = 90°

And PQ ? CD ? ?PQD = 90°

? ?APQ = ?PQD

But they form a pair of alternate angles.

AB ?? CD.

**Q.5.**Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

**Sol.**In the figure, the centre of the circle is 0 and tangent AR touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.

Join OP.

Since tangent at a point to a circle is perpendicular to the radius through that point,

? AB ? OP i.e. ?OPB = 90° …(1)

But by construction,

AB ? PQ ? ?QPB = 90° …(2)

From (1) and (2),

?QPB = ?OPB

which is possible only when O and Q coincide.

Thus, the perpendicular at the point of contact to the tangent passes through the centre.

**Q.6.**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.

**Sol.**?The tangent to a circle is perpendicular to the radius through the point of contact.

?OTA = 90°

Now, in the right ?OTA, we have:

OP

^{2}= OT^{2}+ PT^{2} ? 5

^{2}= OT^{2}+ 4^{2} ? OT

^{2}= 5^{2}– 4^{2} ? OT

^{2}= (5 – 4) (5 + 4) ? OT

^{2}= 1 × 9 = 9 = 3^{2} ? OT = 3

Thus, the radius of the circle is 3 cm.

**Q.7.**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.

**Sol.**In the figure, O is the common centre, of the given concentric circles.

AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.

Since OP is the radius of the smaller circle through P, the point of contact,

? OP ? AB

? ?APB = 90°

Also, a radius perpendicular to a chord bisects the chord.

Now, in right ?APO,

OA

^{2}= AP^{2}– OP^{2} ? 5

^{2}= AP^{2}– 3^{2} ? AP

^{2}= 5^{2}– 3^{2} ? AP

^{2}= (5 – 3) (5 + 3) = 2 × 8 ? AP

^{2}= 16 = (4)^{2} ? AP = 4 cm

Hence, the required length of the chord AB is 8 cm.

**Q.8.**A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:

AB + CD = AD + BC

**Sol.**Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.

AP = AS

BP = BQ

DR = DS

CR = CQ

Adding them, we get

(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)

? AB + CD = BC + DA

which was to be proved.

**Q.9.**In the figure, XY and X’Y’are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and XY’ at B. Prove that ZAOB = W.

**Sol.**?The tangents drawn to a circle from an external point are equal.

? AP = AC

In ? PAO and ? AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

? ?PAO ? ?AOC

? ?PAO = ?CAO

?PAC = 2 ?CAO …(1)

Similarly ?CBQ = 2 ?CBO …(2)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

? ?PAC + ?CBQ = 180°

? 2?CAO + 2 ?CBO = 180° [From (1) and (2)]

? 90° + ?AOB = 180°

? ?AOB = 180° – 90°

? LAOB = 90°.

**Q.10.**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.

**Sol.**Here, let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ? OAP and right ? OBP, we have

PA = PB [Tangents to circle from an external point P]

OA = OB [Radii of the same circle]

OP = OP [Comm]

? By SSS congruency,

? OAP ? OBP

? Their corresponding parts are equal.

?OAA = ?OPB

And ?AOP = ?BOP

? ?APB = 2 ?OPA and ?AOS = 2 ?AOP

But ?AOP = 90° – LOPA

? 2 ?AOP = 180° – 2 ?OPA

? ?AOB = 180° – ?APB

? ?AOB + ?APB = 180°.

**Q.11.**Prove that the parallelogram circumscribing a circle is a rhombus. (CBSE 2012, CBSE Delhi 2014)

**Sol.**We have ABCD, a parallelogram which circumscribes a

circle (i.e., its sides touch the circle) with centre D.

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

? AP = AS

BP = BQ

CR = CQ

DR = DS

Adding, we get

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

? AB + CD = AD + BC

But AB = CD [opposite sides of ABCD]

and BC = AD

? AB + CD = AD + BC ? 2 AB = 2 BC

? AB = BC

Similarly AB = DA and DA = CD

Thus, AB = BC = CD = AD

Hence ABCD is a rhombus.

**Q.12.**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 figure). Find tlic sides AR and AC.

**Sol.**Here ? ABC subscribe the circle with centre O.

Also, radius = 4 cm

? The sides BC, CA and AB touch the circle at D, E and F respectively.

? BF = BD = 8 cm

CE = CD = 6 cm

AF = AE = x cm (say)

? The sides of the triangle are:

14 cm, (x + 6) cm and (x + 8) cm

Perimeter of ? ABC

= [14 + (x + 6) + (x + 8)] cm

= [14 + 6 + 8 + 2x] cm

= 28 + 2x cm

? Semi perimeter of ? ABC

? S – AB = (14 + x) – (8 + x) = 6

S – BC = (14 + x) – (14) = x

S – AC = (14 + x) – (16 + x) = 8

? ar (?ABC) = ar (? OBC) + ar (? OCA) + ar (? OAB)

= 28 cm

^{2}+ (2x + 12) cm^{2}+ (2x + 16) cm^{2} = (28 + 12 + 16) + 4x cm

^{2} = (56 + 4x) cm

^{2} From (1) and (2), we have:

Squaring both sides

(14 + x)2 = (14 + x) 3x

? 196 + x

^{2}+ 28x = 45x + 3x^{2} ? 2x

^{2}+ 14x – 196 = 0 ? x^{2}+ 7x – 98 = 0 ? (x – 7) (x + 14) = 0

? Either x – 7 = 0 ? x = 7

or x +14 = 0 ? x = (–14)

But x = (– 14) is not required

? x = 7 cm

Thus, AB = 8 + 7 = 15 cm

BC = 8 + 6 = 14 cm

CA = 6 + 7 = 13 cm.

**Q.13.**Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Sol.**We have a circle with centre O.

A quadrilateral ABCD is such that the sides AB, BC, CD and DA touch the circle at P, Q,,F R and S respectively.

Let us join OP, OQ, OR and OS. We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

? ?1 = ?2

?3 = ?4

?5 = ?6 and ?7 = ?8

Also, the sum of all the angles around a point is 360°.

? ?1 + ?2 + ?3 + ?4 + ?5 + ?6 + ?7 + ?8 = 360°

? 2 [?1 + ?8 + ?5 + ?4] = 360°

? (?1 + ?8 + ?5 + ?4) = 180° (1)

And 2 [?2 + ?3 + ?6 + ?7] = 360°

? (?2 + ?3) + (?6 + ?7) = 180° (2)

Since, ?2 + ?3 = ?AOB

?6 + ?7 = ?COD

?1 + ?8 = ?AOD

?4 + ?5 = ?BOC

? From (1) and (2), we have:

?AOD + ?BOC = 180° and

?AOB + ?COD = 180°