Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 10 Circles Exercise 10.2

Question 1.

In question 1, 2, 3, choose the correct option and give justification.

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:

A circle where centre is O. The length of the tangent PQ from an external point Q is 24 cm and the distance of Q from centre O is 25 cm.

∴ ∠QPO – 90°

Now, in right triangle QPO,

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

(25)^{2} – (24)^{2} + OP^{2}

or 625 = 576 + OP^{2}

or OP^{2} = 625 – 576

or OP^{2} = 49 = (7)^{2}

or OP = 7 cm.

∴ Option (A) is correct.

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:

In figure OP is the radius and PT is tangent to the circle.

∴ ∠OPT = 90°

Similarly, ∠OQT = 90° and ∠POQ = 110° (Given)

Now POQT is a quadrilateral,

∴ ∠POQ + ∠OQT + ∠QTP + ∠IPO = 360°

⇒ 110° + 90° + ∠QTP + 90° = 360°

or ∠QTP + 290° = 360°

or ∠QTP = 360° – 290?

or ∠QTP = 70°

Option (B) is correct.

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 figure OA is the radius and AP is a tangent to the circle.

∴ ∠OAP = 90°

Similarly, ∠OBP = 90°

Now in right △PAO and △PBO,

∠PAO = ∠PBO = 90°

OP = OP (Common side)

OA = OB

(Radii of the same circle)

∴ ∠PAO = △PBO

[RHS congruence]

∴ ∠AOP = ∠BOP [CPCT]

or ∠AOP = ∠BOP

= \(\frac{1}{2}\)∠AOB ……..(i)

Also, in quadrilateral OAPB,

∠OBP + ∠BPA + ∠PAO + ∠AOB = 360°

90° + 80° + 90° + ∠AOB = 360°

∠AOB = 360° – 260°

∠AOB = 100° …. (ii)

From (i) and (ii),

∠AOP = ∠BOP

= \(\frac{1}{2}\) × 100° = 50°

∴ Option (A) is correct.

Question 4.

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

Solution:

Given : A circle whose centre is O and diameter is AB, l and m are the tangents at points A and B.

To Prove : l || m

Proof : OA is the radius and l is tangent to the circle.

∴ ∠1 = 90°

Similarly, ∠ 2 = 90°

Now, ∠1 = ∠2 = 90°

But there are alternate angles of two parallel lines, when a transversal intersects them.

∴ l || m

Hence, the tangents drawn at the ends of a diameter of acircle and parallel.

(Hence Proved)

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 whose centre is O. PQ is its tangent which meets the circle at A, i.e., the point A is the point of contact of the circle.

To Prove : The perpendicular drawn on the tangent from the point of contact passes through the centre of the circle.

Construction : Join OA.

Proof : Since OA is the radius of the circle and PQ is tangent to the circle where A is the point of contact.

∴ ∠OAP = ∠OAQ = 90°

[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact.]

or OA ⊥ PQ

Since the radius of a circle always passes through the centre of the circle. Therefore, the perpendicular drawn from the point of contact to the tangent passes through the centre of the circle. (Hence Proved)

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:

A circle whose centre is ‘O’. A is any point outside the circle at a distance of 5 cm from its centre.

Length of the tangent = PA

= 4 cm.

Since OP is the radius and PA is tangent to the circle.

∴ ∠OPA = 90°

Now, in right △OPA,

By Pythagoras Theorem,

OA^{2} = OP^{2} + PA^{2}

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

OP^{2} = 25 – 16

OP^{2} = 9 = (3)^{2}

OP = 3 cm.

Hence, 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:

Two concentric circles whose same centre is O and radii are 5 cm and 3 cm respectively.

Let PQ be a chord of the larger circle but tangent to the smaller circle.

Since, OM is the radius of the smaller circle and PMQ is the tangent.

∴ ∠OMP = ∠OMQ = 90°

Now, from right triangles OMP and OMQ,

∠OMP = ∠OMQ = 90°

OP = OQ

[Radii of the same circle]

OM=OM [Common side]

∴ △OMP ≅ △OMQ

[RHS congruence ]

∴ PM = MQ [CPCT]

or PQ = 2 PM = 2 MQ

Now, in right △OMQ,

By Pythagoras Theorem,

OQ^{2} = OM^{2} + MQ^{2}

⇒ (5)^{2} = (3)^{2} + (MQ)^{2}

or MQ^{2} = 25 – 9

or MQ^{2} = 16 = (4)^{2}

or MQ = 4 cm.

∴ Length of chord PQ = 2 QM

= 2 (4) cm.

8 cm.

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

Question 8.

A quadrilateral ABCD is drawn to circumscribe a circle (see Figure). Prove that

AB + CD = AD + BC

OR

In the given figure a quadrilateral PQRS is drawn to circumscribe a circle. Prove that

PQ + RS = PS + QR

Solution:

Given : A quadrilateral ABCD is drawn to circumscribe a circle.

To Prove : AB + CD = AD + BC

Proof : We know that the lengths of the tangents drawn to a circle from an external point are equal.

Now, B is any point outside the circle and BP, BQ are tangent to the circle.

∴ BP = BQ …. (i)

Similarly, AP = AS …. (ii)

and CR= CQ …. (iii)

Also, DR= DS …. (iv)

Adding (i), (ii), (iii) and (iv)

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

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

AB + CD = BC + DA

(Hence Proved)

Question 9.

In 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 ∠AOB = 90°.

Solution:

Given : 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—

To Prove∠AOB = 90°

Construction : Join OC, OA and OB

Proof : We know that the lengths of the tangents drawn from an external point to a circle are equal.

Now, A is any point outside the circle from which two tangents PA and CA are drawn to the circle.

∴ PA = AC

Also, in △PAO and △AOC,

PA = AC (Proved)

OA = OA (Common side)

OP = OC (Radii of the same circle)

∴ △PAO ≅ △AOC [SSS Congruence]

⇒ ∠PAO = ∠CAO [CPCT]

or ∠PAC = 2∠PAO = 2∠CAO …. (i)

Similarly, ∠QBO = ∠COB .

⇒ ∠CBQ = 2∠CBO …. (ii)

Now, ∠PAC + ∠QBC = 90° + 90° = 180°

[∵ OP, OQ are radii and XY, X’Y’ are tangents to the circle]

or 2∠CAO + 2∠OBC = 180°

[Using (i) and (ii)]

or ∠CAO + ∠OBC = \(\frac{180^{\circ}}{2}\) = 90° …. (iii)

Now, in △OAB,

∠CAO + ∠OBC + ∠AOB= 180°

90° + ∠AOB = 180°

[Using (iii)]

or ∠AOB = 180° – 90° = 90°

So, ∠AOB = 90° (Hence Proved)

Question 10.

Prove that the angle between the two tangents drawn from an external point j to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Solution:

Given : A circle where centre is O. From any point P outside die circle PR and PR are tangent lines to the given circle.

To prove : ∠ROQ + ∠QPR = 180°

Proof : OQ is the radius and PQ is the tangent from point P to the given circle.

∴ ∠OQP = 90° … (i)

[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact. ]

Similarly, ∠ORP=90° …. (ii)

V Now, in quadrilateral ROQP,

∠ROQ + ∠PRO + ∠OQP + ∠QPR= 360°

or ∠ROQ + 90° + 90° + ∠QPR = 360° [using equations (i) and (ii)]

or ROQ + ∠QPR + 180° = 360°

or ∠ROQ + ∠QPR = 360° – 180°

or ∠ROQ + ∠QPR = 180°

(Hence Proved)

Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Given : A parallelogram ABCD circumscribing a circle with centre O.

To Prove : ABCD is a rhombus.

Proof : We know that the lengths of the tangents drawn from an external point to a circle are equal.

Now, from any point B outside the circle and BF are two tangents.

∴ BE – BF ….(1)

Similarly, AE = AH ….(2)

and CG = CF ….(3)

Also, DG = DH ….(4)

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

(BE + AE) + (CG + DG)

= (BF + CF) + (AH + DH)

or AB + CD = BC + AD ….(5)

Since it is given that ABCD is a parallelogram

∴ AB = CD and BC = AD ….(6)

From (5) and (6),

AB + AB = BC + BC

or 2AB = 2BC

or AB = BC

Therefore, AB = BC = CD = AD

Therefore ABCD is a rhombus.

(Hence Proved)

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:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm. The sides of the triangle BC, CA, AB touch the circle at points D, E and F respectively. Since the lengths of the tangents drawn from an external point to a circle are equal.

∴ AE = AF = x cm. (say)

CE = CD = 6 cm.

and BF = BD = 8 cm.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ AB; OE ⊥AC and OF ⊥ AB

and OE = OD = OF = 4 cm.

From △ABC,

a = CB = (6 + 8) cm. = 14 cm.

b = AC = (x + 6) cm.

c = BA = (8 + x) cm.

Area of (△BOA) = \(\frac{1}{2}\) × Base × Altitude

= \(\frac{1}{2}\) × (8 + x) × 4

= (16 + 2x) cm^{2} …. (iii)

Area of (△AOC) = \(\frac{1}{2}\) × Base × Altitude

= \(\frac{1}{2}\) × (6 + x) × 4

= (12 + 2x) cm^{2} …. (iv)

ar (△ABC) = ar (△OBC) + ar (△BOA) + ar (△AOC)

\(\sqrt{48 x^{2}+672 x}\) = 28 + 16 + 2x + 12 + 2x

or \(\sqrt{48 x^{2}+672 x}\) = 4x + 56

or \(\sqrt{48 x^{2}+672 x}\) = 4[x + 14]

Squaring both sides

or 48x^{2} + 672x= 16 (x + 14)^{2}

or 48x (x + 14)= 16 (x + 14)^{2}

or 3x = x + 14

or 2x= 14

or x = \(\frac{14}{2}\) = 7

∴ AC= (x + 6) cm.

= (7 + 6) cm = 13 cm

and AB = (x + 8) cm.

= (7 + 8) cm = 15 cm.

So, AB = 15 cm and AC = 13 cm

Question 13.

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

Solution:

Given : A quadrilateral PQRS circumscribing a- circle with centre O where sides PQ, QR, RS and SP touch the circle at L, M, N, T respectively.

To Prove :

∠POQ + ∠SOR = 180°

and ∠SOP + ∠ROQ = 180°

Construction : Join OP, OL, OQ, OM, OR, ON, OS, OT.

Proof : Since tangents drawn from an external point to any circle subtend equal angles at the centre.

∴ ∠2 = ∠3; ∠4 = ∠5; ∠6 = ∠7; ∠8 = ∠1 ……….(i)

We know that the sum of all the angle round a point is 360°.

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

or ∠1 + ∠2 + ∠2 + ∠5 + ∠5 + ∠6 + ∠6 + ∠1 = 360°

or 2(∠ 1 + ∠2 + ∠5 + ∠6) = 360°

or (∠1 + ∠2) + (∠5 + ∠6) = \(\frac{360^{\circ}}{2}\) = 180°

∵ ∠1 + ∠2 = ∠POQ

and ∠5 + ∠6 = ∠SOR

∠POQ + ∠SOR = 180° Similarly, ∠SOP + ∠ROQ = 180°

Hence the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (Hence Proved)

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