## CBSE Class 10 Maths Solution Chapter 10 Circles PDF

**Exercise 10.1**

**Question 1.****How many tangents can a circle have?****Solution:**

There can be infinitely many tangents to a circle.

**Question 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 ……….. .**Solution:****(i)** One**(ii)** Secant**(iii)** Two**(iv)** Point of contact.

**Question 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 PQ is****(a)** 12 cm**(b)** 13 cm**(c)** 8.5 cm**(d)** cm**Solution:**

Radius of the circle = 5 cm

OQ = 12 cm

∠OPQ = 90°

[The tangent to a circle is perpendicular to the radius through the point of contact]

PQ^{2} = OQ^{2} – OP^{2} [By Pythagoras theorem]

PQ^{2} = 12^{2} – 5^{2} = 144 – 25 = 199

PQ = cm.

Hence correct option is (d).

**Question 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.****Solution:**

A line m is parallel to the line n and a line l which is secant is parallel to the given line.

**Exercise 10.2**

**In Q. 1 to 3 choose the correct option and give justification.**

**Question 1.****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**Solution:**

From figure

OQ^{2} = OP^{2} + PQ^{2}

(25)^{2} = OP^{2}+ (24)^{2}

⇒ 625 – 576 = OP^{2}

⇒ 49 = OP^{2}

⇒ OP =

⇒ OP = 7 cm

Radius of the circle = 7 cm

Hence, correct option is **(a)**.

**Question 2.****In figure, if TP and TQ are the two tangents to a circle with centre O so that ****∠POQ = 110°, then ∠****PTQ is equal to**

(a) 60°

(b) 70°

(c) 80°

(d) 90°**Solution:**

∠OPT = 90°

∠OQT = 90°

∠POQ = 110°

TPOQ is a quadrilateral,

∴

∠PTQ + ∠POQ = 180°

⇒

∠PTQ + 110° = 180°

⇒

∠PTQ = 180°- 110° = 70°

Hence, correct option is **(b)**.

**Question 3.****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°**Solution:**

In ∆OAP and ∆OBP

OA = OB [Radii]

PA = PB

[Lengths of tangents from an external point are equal]

OP = OP [Common]

∴ ∆OAP ≅ ∆OBP [SSS congruence rule]

⇒

∠POA = ∠POB = ∠AOB

∠AOB + ∠APB = 180°

⇒

∠AOB + 80° = 180° ……..(i)

⇒

∠AOB = 180°-80° = 100°

From eqn. (i), we get

∠POA = x 100° = 50°

Hence, correct option is **(a)**.

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

AB is a diameter of the circle, p and q are two tangents.

OA ⊥ p and OB ⊥ q

∴

∠1 = ∠2 = 90°

⇒ p ∥q [ ∠1 and ∠2 are alternate angles]

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

XY tangent to the circle C(0, r) at B and AB ⊥ XY. Join OB.

∠ABY = 90° [Given]

∠OBY = 90°

[Radius through point of contact is perpendicular to the tangent]

∴

∠ABY + ∠OBY = 180°

∴

⇒ ABO is collinear

∴ AB passes through centre of the circle.

**Question 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.****Solution:**

OA = 5 cm, AP = 4 cm

OP = Radius of the circle

∠OPA = 90° [Radius and tangent are perpendicular]

**Question 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.****Solution:**

Radius of larger circle = 5 cm

Radius of smaller circle = 3 cm

OP ⊥ AB

[Radius of circle is perpendicular to the tangent]

AB is a chord of the larger circle

∴ OP bisects AB

**Question 8.****A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.****Solution:**

AP = AS …(i)

[Lengths of tangents from an external point are equal]

BP = BQ …(ii)

CR = CQ …(iii)

DR = DS …(iv)

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

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

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

⇒ AB + CD = AD + BC

Hence proved.

**Question 9.****In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ****∠****AOB = 90°.****Solution:****Given:** Two parallel tangents to a circle with centre O.

Tangent AB with point of contact C intersects XY at A and X’Y’ at B**To Prove:** ∠AOB = 90°**Construction:** Join OA, OB and OC**Proof:** In ∆AOP and ∆AOC

AP = AC [ Lengths of tangents]

OP = OP [ Radii]

OA = OA [ Common]

⇒ ∆AOP ≅ ∆AOC [SSS congruence rule]

⇒

∠PAO = ∠CAO [C.P.C.T]

∴

∠PAC = 2∠OAC ….(i)

Similarly ∠QBC = 2 ∠OBC ….(ii)

Adding (i) and (ii), we get

∠PAC + ∠QBC = 2[∠OAC + ∠OBC]

∵

∠PAC + ∠QBC = 180°

[interior consecutive angle on same side of transversal]

∴ 180° = 2[∠OAC + ∠OBC]

⇒

∠OAC + ∠OBC = 90°

In ∆AOB,

∠AOB + [∠OAC + ∠OBC] = 180°

⇒

∠AOB + 90° = 180°

⇒

∠AOB = 180° – 90° = 90°

**Question 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.****Solution:**

PA and PB are two tangents, A and B are the points of contact of the tangents.

OA ⊥ AP and OB ⊥BP

∠OAP = ∠OBP = 90°

[Radius and tangent are perpendicular to each other]

p In quadrilateral OAPB

∠OAP + ∠OBP + ∠APB + ∠AOB = 360°

⇒ 90° + 90° + ∠APB + ∠AOB = 360°

∠APB + ∠AOB = 360°- 180° = 180°

Hence, ∠APB and ∠AOB are supplementary angles.

**Question 11.****Prove that the parallelogram circumscribing a circle is a rhombus.****Solution:**

Parallelogram ABCD circumscribing a circle with centre O.

OP ⊥ AB and OS ⊥AD

In ∆OPB and ∆OSD

∠OPB = ∠OSD [Each 90°]

OB = OD [Diagonals of a parallelogram bisect each other]

OP = OS

⇒ ∆OPB ≅ ∆OSD [ RHS congruence rule]

PB = SD ….. (i) [C.P.C.T]

AP = AS …….(ii) [ Lengths of tangents]

Adding (i) and (ii), AP + PB = AS + DS

AB = AD

Similarly AB = BC = CD = DA

Hence, parallelogram ABCD is a rhombus.

**Question 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 the sides AB and AC.****Solution:**

BD = 8 cm and DC = 6 cm

BE = BD = 8 cm

CD = CF = 6 cm

Let AE = AF = x cm

In ∆ABC, a = 6 + 8 = 14 cm

b = (x + 6) cm

c = (x + 8) cm

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

AB touches at P.

BC, CD and DA touch the circle at Q, R and S.**Construction:** Join OA, OB, OC, OD and OP, OQ, OR, OS.

OA bisects ∠POS

∴

∠1 = ∠2

Similarly ∠3 = ∠4

∠5 = ∠6

∠7 = ∠8

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

∠1 + ∠1 + ∠4 + ∠4. + ∠5 + ∠5 + ∠8 + ∠8 = 360°

2[∠1 + ∠4 + ∠5 + ∠8] = 360°

(∠1 + ∠8 + ∠4 + ∠5) = 180°

∠AOD + ∠BOC = 180°

Similarly, ∠AOB + ∠COD = 180°

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

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