**KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2

In Questions 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:

A) 7 cm.

In ⊥∆OPQ, ∠P = 90°

∴ As per Pythagoras theorem,

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

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

OP^{2} + 576 = 625

OP^{2} = 625 – 576

OP^{2} = 49

∴ OP = 7 cm.

Question 2.

In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

Solution:

(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°

∴ OP⊥PT and OQ⊥QT

⇒ ∠OPT = 90° and ∠OQT = 90°

Now, in the quadrilateral TPOQ, we get

∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]

⇒ ∠PTQ + 290° = 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) 50°

OA, OB are radii,

PA, PB are tangents

∴ ∠PAO = 90°, ∠PBO = 90°

In quadrilateral OAPB,

∠BOA = 180° – 80° = 100°

But

∴∠POA = 50°.

Question 4.

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

Solution:

In the figure, 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 tangents 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

Hence, the two tangents are parallel.

Question 5.

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

Solution:

Data: Perpendicular at the point of contact to the tangent to a circle passes through the centre.

PQ is the tangent to circle with centre ’O’.

Let the perpendicular drawn at the point P is ∠RPQ.

∠RPQ = 90° …………. (i)

Radius drawn to circle at the point of contact is perpendicular.

∴ ∠OPQ = 90° …………. (ii)

From eqn. (i) and eqn. (ii), we have

∠RPQ = ∠OPQ = 90° .

This is a contradiction, because

∠RPQ is the part of ∠OPQ.

∴ ∠RPQ < ∠OPQ

∴ ∠RPQ ≠∠OPQ.

∴ The perpendicular at the point of contact to the tangent to a circle passes through the centre.

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:

Data: OB is the radius, BA is the tangent.

∴ ∠OBA = 90°, OA = 5 cm. (data)

Length of tangent, AB = 4 cm.

∴ Radius, OB =?

In ⊥∆OBA, ∠OBA = 90°

∴ OB^{2} + BA^{2} = OA^{2} .

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

OB^{2} + 16 = 25

OB^{2} = 25 – 16

OB^{2} = 9

∴ OB = 3 cm.

∴ Radius of circle, OB = 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:

Radius of big circle,

OA = OB = 5 cm.

Radius of small circle, OP = 3 cm.

Angle between radius and tangent is 90°.

∴ ∠OPA = ∠OPB = 90°

( ∵ Chord AB is tangent to small circle.)

Now, in ⊥∆ OPA, ∠OPA = 90°

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

(3)^{2} + AP^{2} = (5)^{2}

9 + AP^{2} = 25

∴ AP^{2} = 25 – 9

AP^{2} = 16

∴ AP = 4 cm.

Similarly, in ⊥∆OPB, PB = 4 cm.

∴ Length of chord, AB = AP + PB = 4 + 4

∴ Chord, AB = 8 cm.

Question 8.

A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC

Solution:

Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches 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 and CR = CQ

Adding them, we get

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

⇒ AB + CD = BC + DA

Question 9.

In the following figure, 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°.

Solution:

Data: 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°.

Tangent XY || Tangent X’Y’

∴ ∠PAB + ∠QBA = 180° (∵ interior angles)

∠OAB + ∠OBA = 90°

Now, in AOAB,

∠AOB + ∠OAB + ∠OBA = 180°

∠AOB + 90° = 180°

∴ ∠AOB = 180° – 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-segment joining the points of contact at the centre.

Solution:

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]

OA = OB [Radii of the same circle]

OP = OP [Common]

⇒ ∆OAP ≅ ∆OBP [By SSS congruency]

∴ ∠OPA = ∠OPB [By CPCT]

and ∠AOP = ∠BOP

⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP

In right ∆OAP,

∠AOP + ∠OPA + ∠PAO = 180°

⇒ ∠AOP = 180° – 90° – ∠OPA

⇒ ∠AOP = 90° – ∠OPA

⇒ 2∠AOP = 180° – 2∠OPA

⇒ ∠AOB = 180° – ∠APB

⇒ ∠AOB + ∠APB = 180°

Question 11.

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

To prove: Parallelogram ABCD is a rhombus.

ABCD is a parallelogram in which ‘O’ is the centre of the interior circle drawn in the parallelogram.

Tangents AC and BD are drawn which intersect at ‘OP’.

Sides of the parallelogram AB. BC. CD and AD touch at the points P, Q, R, and S respectively.

OP, OS joined.

We have OP ⊥ AB, OS ⊥ AD.

In ⊥∆ OPB and ∆OSD,

∠OPB = ∠OSD =90°

OB = OD (tangent is bisected)

OP = OS (Radii)

∴ ∆OPB = ∆OSD

∴ PB = SD → (1)

AP = AS → (2) (∵ tangent of the external point)

By combining eqn. (1) and Eqn. (2),

AP + PB = AS + SC

AB = AD

Similarly, AB = BC = CD = DA

∴ 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 the figure given). Find the sides AB and AC.

Solution:

To Prove: Side AB = ? and Side AC = ?

BD = 8 cm. DC = 6 cm.

BE = 8 cm. CF = 6 cm.

AE = AF = x cm.

In ∆ABC,

a = BC = BD + DC =8 + 6 =14 cm.

b = CA = (x + 6) cm.

c = AB = (x + 8) cm.

\(S=\frac{a+b+c}{2} \quad=\frac{14+(x+6)+(x+8)}{2}\)

\(=\frac{2 x+28}{2}\)

= (x + 14) cm.

Area of ∆ABC = \(=\sqrt{s(s-a)(s-b)(s-c)}\)

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

\(=\sqrt{48 x \times(x+14)} c m^{2}\) …………….. (i)

Area of ∆ABC

= ∆OBC + ∆OCA + ∆OAB W.

\(=\frac{1}{2} \times 4 \times a+\frac{1}{2} \times 4 \times b+\frac{1}{2} \times 4 \times c\)

– 2(a + b + c)

= 2 × 2S

= 4S

= 4(x +14) cm^{2}………………. (ii)

From eqn. (i) and eqn. (ii),

\(\sqrt{48 x \times(x+14)}=4(x+14)\)

48x × (x + 14) = 16 x (x + 14)2

3x = x + 14

∴ x = 7 cm.

AB = c = x + 8 = 7 + 8 = 15 cm.

AC = b = x + 6 = 7 + 6 = 13 cm.

Question 13.

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

Solution:

We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.

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 and ∠4 + ∠5 = ∠BOC

∴ From (1) and (2), we have

∠AOD + ∠BOC = 180°

and ∠AOB + ∠COD = 180°

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2, drop a comment below and we will get back to you at the earliest.