Question 1

How many tangents can a circle have?

Solution: A circle has infinitely many points on its perimeter. Since a tangent can be drawn touching the circle at each point, a circle can have infinitely many tangents.

Question 2

Fill in the blanks:

1. A tangent to a circle intersects it in one point(s).
2. A line intersecting a circle in two points is called a secant.
3. A circle can have two parallel tangents at the most.
4. The common point of a tangent to a circle and the circle is called the 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) √119 cm

Solution:
Radius OP = 5 cm, OQ = 12 cm.
By Theorem 10.1, OP ⊥ PQ. Therefore, ΔOPQ is a right-angled triangle at P.
By Pythagoras Theorem:
OQ² = OP² + PQ²
12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 144 - 25 = 119
PQ = √119 cm.
Correct Option: (D) √119 cm

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:
1. Draw a circle with centre O.
2. Draw a given line L.
3. Draw line A parallel to L touching the circle at one point (Tangent).
4. Draw line B parallel to L intersecting the circle at two points (Secant).

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:
Let O be the centre, OP be radius, PQ = 24 cm, OQ = 25 cm.
OP ⊥ PQ (Theorem 10.1)
In right ΔOPQ:
OQ² = OP² + PQ²
25² = OP² + 24²
625 = OP² + 576
OP² = 625 - 576 = 49
OP = √49 = 7 cm.
Correct Option: (A) 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:
OP ⊥ PT and OQ ⊥ QT, so ∠OPT = 90° and ∠OQT = 90°.
In quadrilateral OPTQ, sum of interior angles = 360°:
∠OPT + ∠POQ + ∠OQT + ∠PTQ = 360°
90° + 110° + 90° + ∠PTQ = 360°
290° + ∠PTQ = 360°
∠PTQ = 360° - 290° = 70°.
Correct Option: (B) 70°

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 quadrilateral AOBP:
∠OAP = 90°, ∠OBP = 90°, ∠APB = 80°.
∠AOB = 360° - (90° + 90° + 80°) = 100°.
Line OP bisects ∠AOB, so ∠POA = ∠AOB / 2 = 100° / 2 = 50°.
Correct Option: (A) 50°

Question 4

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

Solution:
Let AB be a diameter of a circle with centre O.
Let PQ be the tangent at A and RS be the tangent at B.
Since radius is perpendicular to tangent:
∠OAP = 90° and ∠OBS = 90°.
Thus, ∠BAP = 90° and ∠ABS = 90°.
Since alternate interior angles ∠BAP and ∠ABS are equal (∠BAP = ∠ABS = 90°), the line PQ is parallel to RS (PQ || RS). Hence proved.

Question 5

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

Solution:
Let AB be the tangent at point P to a circle with centre O.
Suppose perpendicular to AB at P does not pass through O; let it pass through another point O'.
Then ∠O'PB = 90°.
By Theorem 10.1, OP ⊥ AB, so ∠OPB = 90°.
Comparing both: ∠O'PB = ∠OPB = 90°.
This is possible only if O' coincides with O. Hence, the perpendicular at the point of contact 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:
Let O be centre, P be point of contact.
OA = 5 cm, AP = 4 cm, OP = radius r.
OP ⊥ AP, so in right ΔOPA:
OA² = OP² + AP²
5² = r² + 4²
25 = r² + 16
r² = 9 ⇒ r = 3 cm.
Radius of the circle = 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 AB be chord of larger circle touching smaller circle at P. Centre is O.
Radius of larger circle OA = 5 cm, radius of smaller circle OP = 3 cm.
OP ⊥ AB, and perpendicular from centre to chord bisects the chord (AP = PB).
In right ΔOPA:
OA² = OP² + AP²
5² = 3² + AP²
25 = 9 + AP² ⇒ AP² = 16 ⇒ AP = 4 cm.
Length of chord AB = 2 × AP = 2 × 4 = 8 cm.

Question 8

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

Solution:
Let circle touch sides AB, BC, CD, DA at P, Q, R, S respectively.
By Theorem 10.2 (tangents from external point are equal):
AP = AS   ---(1)
BP = BQ   ---(2)
CR = CQ   ---(3)
DR = DS   ---(4)
Adding equations (1), (2), (3), (4):
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC. Hence proved.

Question 9

In given 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:
Join OC, OA, OB.
ΔOPA ≅ ΔOCA (by SSS congruence rule)
So, ∠POA = ∠COA ⇒ ∠POC = 2∠COA.
Similarly, ΔOQB ≅ ΔOCB ⇒ ∠QOB = ∠COB ⇒ ∠QOC = 2∠COB.
Since POQ is a straight line diameter, ∠POC + ∠QOC = 180°.
2∠COA + 2∠COB = 180°
2(∠COA + ∠COB) = 180°
∠AOB = 90°. Hence proved.

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 tangents from P with point of contact A and B. Centre is O.
∠OAP = 90° and ∠OBP = 90°.
In quadrilateral OAPB:
∠APB + ∠AOB + ∠OAP + ∠OBP = 360°
∠APB + ∠AOB + 90° + 90° = 360°
∠APB + ∠AOB = 180°.
Therefore, angles are supplementary. Hence proved.

Question 11

Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:
Let ABCD be a parallelogram circumscribing a circle.
Since ABCD is a parallelogram, AB = CD and AD = BC.
From Question 8, for circumscribed quadrilateral: AB + CD = AD + BC.
Substituting AB = CD and AD = BC:
AB + AB = AD + AD
2 AB = 2 AD ⇒ AB = AD.
Since adjacent sides are equal, parallelogram 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. Find the sides AB and AC.

Solution:
Let circle touch AB at E and AC at F.
BD = BE = 8 cm, CD = CF = 6 cm, AE = AF = x cm.
Sides of ΔABC: a = 14 cm, b = (x + 6) cm, c = (x + 8) cm.
Semi-perimeter s = (14 + x + 6 + x + 8) / 2 = x + 14.
Area by Heron's formula = √[s(s-a)(s-b)(s-c)] = √[(x + 14)(x)(8)(6)] = √[48x(x + 14)].
Area by triangles = Area(ΔOBC) + Area(ΔOCA) + Area(ΔOAB)
= (1/2 × 14 × 4) + (1/2 × (x+6) × 4) + (1/2 × (x+8) × 4)
= 28 + 2x + 12 + 2x + 16 = 4(x + 14).
Equating both area formulas:
√[48x(x + 14)] = 4(x + 14)
Squaring both sides: 48x(x + 14) = 16(x + 14)²
3x = x + 14 ⇒ 2x = 14 ⇒ x = 7 cm.
Thus, AB = x + 8 = 7 + 8 = 15 cm.
AC = 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:
Let ABCD circumscribe a circle touching at P, Q, R, S. Join vertices and points of contact to center O.
ΔAOP ≅ ΔAOS ⇒ ∠1 = ∠8.
Similarly, ∠2 = ∠3, ∠4 = ∠5, ∠6 = ∠7.
Sum of all angles at O = 360°:
(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°
2∠1 + 2∠2 + 2∠5 + 2∠6 = 360°
∠1 + ∠2 + ∠5 + ∠6 = 180°
(∠1 + ∠2) + (∠5 + ∠6) = 180°
∠AOB + ∠COD = 180°.
Similarly, ∠AOD + ∠BOC = 180°. Hence proved.