Gujarat Board GSEB Solutions Class 10 Maths Chapter 10 Circles Ex 10.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 10 Circles Ex 10.2

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) 12cm

(c) 15 cm

(d) 24.5 cm

Solution:

(a) From the Fig.

QP = 24cm

QO = 25cm .

Also, OP ⊥ PQ

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

In right ΔOPQ, by Pythagoras theorem

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

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

OP^{2} = 25^{2} – 24^{2}

OP^{2} = 625 – 576 = 49

OP = 7 cm

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:

(b) ∠OPT = ∠OQT = 90°

[Tangents perpendicular to the radius

through the point of contact]

In quadrilateral OPTQ

∠OPT + ∠PTQ + ∠OQT + ∠POQ = 360°

[Sum of all angles of quadrilateral]

⇒ 90° + ∠PTQ + 90° + 110° = 360°

290° + ∠PTQ = 360°

⇒ ∠PTQ = 360° – 290° = 70°

Question 3.

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

(a) 50°

(b) 60°

(c) 70°

(d) 80°

Solution:

(a) Join OP

In ΔPOA and ΔPOB

PA = PB [Tangents from external point P]

OA = OB

[Radii of same circle Common]

OP = OP

⇒ ΔPOA ≅ ΔPOB [SSS congruency rule]

⇒ ∠OPA = ∠OPB = 40°

Now, in ΔOAP

∠POA + ∠OAP + ∠APO = 180°

[Angle sum property of triangle]

∠POA + 90° + 40° = 180°

∠POA + 130° = 180°

∠POA = 50°

Question 4.

Prove that the tangent lines at the ends of a diameter of a circle are parallel.

Solution:

From Fig.

∠1 = 90° ……….(1)

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

∠2 = 90° ……….(2)

From (1) and (2),

∠1= ∠2

But these are alternate interior angles.

∴ PQ || RS

Question 5.

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

Solution:

Given:

A circle with centre O.

AB is a tangent at point A.

O’A ⊥ AB

To prove: AO’ passes through the centre O.

Construction: Join OA.

Proof: ∠O’AB = 90° (Given)

∠OAB = 90°

[Radius is perpendicular at the point of contact]

⇒ ∠O’AB = ∠OAB = 90°

which is possible only when O’ and O coincides

with each other.

⇒ AO’ and AO is the same line.

∴ AO’ passes through the centre O.

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:

∠OPA = 90° [Theorem 10.1]

∴ OA^{2} = OP^{2} + AP^{2} [By Pythagoras theorem]

5^{2} = OP^{2} + 4^{2}

OP^{2} = 25 – 16 = 9

OP = \(\sqrt{9}\) = 3 cm

∴ The radius of the circle is 3 cm.

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:

Let O be the common centre of two concentric circles.

AB, the chord of the larger circle which touches the smaller circle at M.

Join 0M and OA.

Then ∠OMA = 90° [Theorem 10.11]

∴ OMA is a right-angled triangle.

OA^{2} = OM^{2} + AM^{2} [By Pythagoras theorem]

5^{2} + 3^{2} + AM^{2}

25 = 9 + AM^{2}

AM^{2} = 25 – 9 = 16

AM = \(\sqrt{16}\) = 4cm

Now, AM = BM = 4 cm

[The perpendicular from the centre of a circle to a chord bisects the chord]

∴ AB = AM + BM

= AM + AM = 2AM

2 x 4 = 8 cm

Question 8.

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

[CBSE 2008, 09, 12, 13, 16, 171

Solution:

Since the tangents drawn from an external point are equal in length.

∴ AP = AS ………(1)

BP = BQ ……..(2)

CR = CQ ………(3)

DR = DS ………(4)

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

(AP + BP) + (CR + DR)

⇒ AS + BQ + CQ + DS

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Question 9.

In Fig. XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°. [CBSE 2011, 2012, 2013, 2017]

Solution:

Given: XY and X’Y’ are two parallel tangents with centre O and AB is another tangent at the point of contact C and intersecting XY at A and X’Y’ at B.

To prove: ∠AOB = 90°

Construction: Join OC.

Proof: ∠OPA = 90° [Theorem 10.11]

∠OCA = 90°

∴ In ΔOPA and ΔOCA

OA = OA (Common)

AP = AC [Tangents drawn from an external point are equal in length ]

and ∠OPA = ∠OCA = 90°

⇒ ΔSOPA ≅ ΔSOCA [RHS congruence criteria]

⇒ ∠OAP = ∠OAC [CPCT]

⇒ ∠OAC = \(\frac {1}{2}\) ∠PAB

Similarly, ∠OBQ = ∠OBC

⇒ ∠OBC = ∠QBA ……….(2)

∴ XY || XY,

and a transversal AB intersects them

⇒ ∠PAB + ∠QBA = 180° [Sum of the cointerior angles is 180° ]

⇒ \(\frac {1}{2}\) ∠PAB + \(\frac {1}{2}\)∠QBA = \(\frac {1}{2}\) x 180°

∠PAB + ∠QBA = 90° ………(3)

[From (1) and (2)]

In ΔAOB

∠OAC + ∠OBC + ∠AOB = 180°

[Angle sum property of triangle]

⇒ 90° + ∠AOB = 180° [From (3)]

∠AOB = 180° – 90° = 90°

∠AOB = 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 segments joining the points of contact at the centre. [CBSE 2012]

Solution:

From Fig.

∠OPA = 90° ……..(1)

∠OBP = 90 ……….(2)

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

Also, OAPB is a quadrilateral.

∴ ∠APB + ∠OBP + ∠AOB + ∠OAP = 360°

[Angle sum property of quadrilateral]

∠APB + 90° + ∠AOB + 90° = 360°

[From (1) and (2)]

∠APB + ∠AOB = 360° – 180° = 180°

∠APB and ∠AOB are supplementary angles.

Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus. [CBSE 20121

Solution:

Given: ABCD is a parallelogram.

To prove: ABCD is a rhombus.

Proof: The tangents drawn from an external point are equal in length.

∴ AP = AS …….(1)

PB = BQ ………(2)

CR = CQ ……..(3)

DR = DS ……..(4)

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

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

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

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC

(Opposite sides of parallelogram are equal]

⇒ 2AB = 2BC

⇒ AB = BC

Now AD = BC = AB = CD

⇒ 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:

Given CD 6 cm

BD = 8 cm and radius = 4 cm

Join OC, OA and OB.

As the tangents drawn from an external point are equal in length

∴ CD = CF = 6cm

BD = BE = 8 cm

and let AF = AE = x cm

In ΔOCB,

Area of ΔOCB

= \(\frac {1}{2}\) base x height

= \(\frac {1}{2}\) x CB x OD

= \(\frac {1}{2}\) x 14 x 4 = 28 cm2

Similarly, Area of ΔOCA

= \(\frac {1}{2}\) x AC x OF

= \(\frac {1}{2}\) x (6 + x) x 4

= (12 + 2x) cm^{2}

Similarly, Area of ΔOBA

= \(\frac {1}{2}\) x AB x OE

= (8 + x) x 4

=(16 + 2x) cm^{2}

Now, area of ΔABC

= ar(OCB) + ar(OCA) + ar(OBA)

= 28 + (12 + 2x) + (16 + 2x)

= 56 + 4x …………(1)

Now, perimeter of ΔABC

= AB + BC + CA

= x + 6 + 14 + 8 + x

= 28 + 2x

S = \(\frac {1}{2}\) (28 + 2x) = 14 + x

By Heron’s formula,

area of ΔABC = \(\sqrt{S(S-a)(S-b)(S-c)}\)

= \(\sqrt{(14 + x)(14 + x – 14)(14 + x – x – 6)(14 + x – x – 8)}\)

= \(\sqrt{(14 + x) \cdot x \cdot(8)(6)}\)

= \(4 \sqrt{3 x(14 + x)}\) …….(2)

From equation (1) and (2), we get

56 + 4x = \(4 \sqrt{3 x (14 + x)}\)

14 + x = \(\sqrt{3 x(14 + x)}\)

On squaring both sides,

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

14 + x = 3x

2x = 14

x = 7

∴ Length AC = 6 + x = 6 + 7 = 13cm

Length AB = 8 + x = 8 + 7 = 15 cm

Question 13.

Prove that the opposite sides of a quadrilateral circumscribing a circle subtended supplementary angles of the centre of the circle. [CBSE 2012, 2017]

Solution:

Given:

ABCD is a quadrilateral circumscribing a circle whose centre is O.

To prove:

(i) ∠BOC + ∠AOD = 180°

(ii) ∠AOB + ∠COD = 180°

Construction. Join OP, OQ, OR and OS;

where P, Q, R and S are points of contacts of tangents AB, BC, CD and DA to the circle.

Proof:

(i) As tangents drawn from an external point are equal in length.

In ΔOBP and ΔOBQ

OP = OQ [Radii of the same circle]

OB = OB [Common]

BP = BQ [From(1)]

∴ ΔOBP ≅ ΔOBQ [SSS congruency rule]

∴ ∠5 = ∠6 [CPCT]

Similarly,

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

[Sum angle property of quadrilateral]

= ∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360° [From (2)]

2[∠1 + ∠4 + ∠5+ ∠8] = 360°

∠1 + ∠4 + ∠5 + ∠8 = 180°

(∠1 + ∠8) + (∠4 + ∠5) = 180°

∠AOD + ∠BOC = 180°

(ii) Similarly, we can prove that

∠AOB + ∠COD = 180°