1. The distance between two parallel tangents of a circle of radius 4 cm is

(a) 2 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

Answer: d

Explanation:

Reason: Here radius, r = 4 cm

Required distance,

AB = OA + OB

= r + r = 2r = 2×4 = 8 cm

2. In the given figure, if ∠RPS = 25°, the value of ∠ROS is

(a) 135°

(b) 145°

(c) 165°

(d) 155°

Answer: d

Explanation: Reason: Since OR ⊥ PR and OS ⊥ PS

∴ ∠ORP = ∠OSP = 90°

In □ ORPS, ∠ROS + ∠ORP + ∠RPS + ∠OSP = 360°

∠ROS + 90° + 25° + 90° = 360°

∠ROS = 360° – 205° = 155°

3. A tangent is drawn from a point at a distance of 17 cm of circle C(0, r) of radius 8 cm. The length of its tangent is

(a) 5 cm

(b) 9 cm

(c) 15 cm

(d) 23 cm

Answer: c

Explanation:

Reason: In right angle ∆OAP, AP² + OA² = OP²

⇒ AP² + (8)² = (17)² => AP² + 64 = 289

⇒ AP² = 289 – 64 = 225

∴ AP = √225 = 15 cm

4. The length of tangents drawn from an external point to the circle

(a) are equal

(b) are not equal

(c) sometimes are equal

(d) are not defined

Answer: a

Explanation: Reason: Since the length of tangents drawn from an external point to a circle are equal.

5. Number of tangents drawn at a point of the , circle is/are

(a) one

(b) two

(c) none

(d) infinite

Answer: a

Explanation: Reason: There is only one tangent at a point of the circle.

6. The tangents drawn at the extremities of the diameter of a circle are

(a) perpendicular

(b) parallel

(c) equal

(d) none of these

Answer: b

Explanation:

Reason: Since OP ⊥ AB and OQ ⊥ CD

∴ Z1 = 90° and Z2 = 90°

⇒ ∠1 = Z2, which are alternate angles.

∴ AB || CD

7. Tangents from an external point to a circle are

(a) equal

(b) not equal

(c) parallel

(d) perpendicular

Answer: a

Explanation: Reason: Tangents from external points to a circle are equal.

8. The length of a tangent drawn from a point at a distance of 10 cm of circle is 8 cm. The radius of the circle is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 7 cm

Answer: c

Explanation:

Reason: In right angle ΔOAP, we have

OA² + AP² = OP²

⇒ OA² + (8)² = (10)2

⇒ OA² + 64 = 100

⇒ OA² = 100 – 64 = 36

∴ OA = √36 = 6 cm

9. In given figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 6 cm then the length of BR is

(a) 6 cm

(b) 5 cm

(c) 4 cm

(d) 3 cm

Answer: b

Explanation: Reason: Since

BQ = BR …(i) [∵ Tangents drawn from external points are equal]

CQ = CP …[Using (i)]

BC + BQ = 11

⇒ 6 + BR = 11

⇒ BR = 11 – 6 = 5 cm

10. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(a) 60 cm²

(b) 65 cm²

(c) 30 cm²

(d) 32.5 cm²

Answer: a

Explanation:

Reason: OP² = OQ² + PQ²

169 = 25 + PQ²

PQ² = 144

PQ = 12

Area PQOR = ar (AOPQ) + ar (AOPR)

= 1/2 × 12 × 5 + 1/2 × 12 × 5 = 60 cm²

11. In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°, then ∠BOC is equal to [AI 2011]

(a) 40°

(b) 50°

(c) 140°

(d) 150°

Answer: c

Explanation:

(c) In quadrilateral ABOC

∠ABO + ∠BOC + ∠OCA + ∠BAC = 360°

⇒ 90° + ∠BOC + 90° + 40° = 360°

⇒ ∠BOC = 360° – 220° = 140°

12. In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is [Foreign 2011]

(a) 25 cm

(b) 26 cm

(c) 24 cm

(d) 10 cm

Answer: d

Explanation: (d) v OT is radius and PT is tangent

∴ OT ⊥ PT

Now, in AOTP,

⇒ OP² = PT² + OT²

⇒ 26² = 24² + OT²

⇒ 676 – 576 = OT²

⇒ 100 = OT²

⇒ 10 cm = OT

13. A line through point of contact and passing through centre of circle is known as

(a) tangent

(b) chord

(c) normal

(d) segment

Answer: c

Explanation: (c) normal

14. In figure if O is centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to [NCERT Exemplar Problems]

(a) 100°

(b) 80°

(c) 90°

(d) 75°

Answer: a

Explanation:

(a) OP ⊥ PR [Y Tangent and radius are ⊥ to each other at the point of contact]

∠OPQ = 90° – 50° = 40°

OP = OQ [Radii]

∴ ∠OPQ = ∠OQP = 40°

In ∆OPQ,

⇒ ∠POQ + ∠OPQ + ∠OQP = 180°

⇒ ∠POQ + 40° + 40° = 180°

∠POQ = 180° – 80° = 100°.

15. In figure AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to [NCERT Exemplar Problems]

(a) 4 cm

(b) 2 cm

(c) 2√3 cm

(d) 4√3 cm

Answer: c

Explanation:

16. Two parallel lines touch the circle at

points A and B respectively. If area of the circle is 25 n cm2, then AB is equal to

(a) 5 cm

(b) 8 cm

(c) 10 cm

(d) 25 cm

Answer: c

Explanation:

(c) Let radius of circle = R

∴ πR² = 25π

⇒ R = 5 cm

∴ Distance between two parallel tangents = diameter = 2 × 5 = 10 cm.

17. C (O, r_{1}) and C(O, r_{2}) are two concentric circles with r_{1} > r_{2} AB is a chord of C(O, r_{1}) touching C(O, r,_{2}) at C then

(a) AB = r_{1}

(b) AB = r_{2}

(c) AC = BC

(d) AB = r_{1} + r_{2}

Answer: c

Explanation:

(c) ∵ AB touches

C(0, r<sub>2</sub>)

∴ OC ⊥ AB

Also, perpendicular from the centre to a chord bisects the chord.

∴ AC = BC

18. In figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then ∠BAT is equal to [Delhi 2011]

(a) 100°

(b) 40°

(c) 50°

(d) 90°

Answer: c

Explanation:

(c) ∠AOB = 100°

∠OAB = ∠OBA (∵ OA and OB are radii)

Now, in ∆AOB,

∠AOB + ∠OAB + ∠OBA = 180°

(Angle sum property of A)

⇒ 100° + x + x = 180° [Let ∠OAB = ∠OBA = x]

⇒ 2x = 180° – 100°

⇒ 2x = 80°

⇒ x = 40°

Also, ∠OAB + ∠BAT = 90°

[∵ OA is radius and TA is tangent at A]

⇒ 40° + ∠ BAT = 90°

⇒ ∠BAT = 50°

19. In the figure PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then ∠OAB is [Delhi 2011]

(a) 30°

(b) 60°

(c) 90°

(d) 15°

Answer: a

Explanation:

(a) Given ∠APB = 60°

∵ ∠APB + ∠PAB + ∠PBA = 180°

⇒ APB + x + x = 180°

[∵ PA = PB ∴ ∠PAB = ∠PBA = x (say)]

⇒ 60° + 2x = 180°

⇒ 2x = 180° – 60°

⇒ 2x = 120°

⇒ x = 120°/2 = 60°

Also, ∠OAP = 90°

⇒ ∠OAB + ∠PAB = 90°

⇒ ∠OAB + 60°= 90°

⇒ ∠OAB = 30°

20. In the given figure, TP and TQ are two tangents to a circle with centre O, such that ∠POQ = 110°. Then ∠PTQ is equal to [Foreign 2011]

(a) 55°

(b) 70°

(c) 110°

(d) 90°

Answer: b

Explanation:

(b) In quadrilateral POQT,

∠PTQ + ∠TPO + ∠TQO + ∠POQ

= 360°

⇒ ∠PTQ + 90° + 90° + 110° = 360°

⇒ ∠PTQ + 290° = 360°

⇒ ∠PTQ = 360° – 290° = 70°

21. In figure,PQ and PR are tangents to a circle with centre A. If ∠QPA=27°, then ∠QAR equals to [Foreign 2012]

(a) 63°

(b) 153°

(c) 110°

(d) 90°

Answer: c

Explanation:

(c) ∠QPA = ∠RPA

[∵ ∆AQP ≅ ∆ARP (RHS congruence rule)]

⇒ ∠RPA = 27°

∴ ∠QPR = ∠QPA + ∠RPA

= 27° + 27° = 54°

Now,

∠QAR + ∠AQP + ∠ARP + ∠QPR = 360°

⇒ ∠QAR = 90° + 90° + 54° = 360°

⇒ ∠QAR = 360° – 234° = 126°

22. In figure if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar Problems]

(a) 20°

(b) 40°

(c) 35°

(d) 45°

Answer:

Explanation:

(b) AB || PR

∴ ∠ABQ =∠BQR

[Alternate interior angles]

⇒ ∠ABQ = 70°

Also, ∠BQR = ∠BAQ [Angles in alternate segment]

⇒ ∠BAQ = 70°

In ∆AQB,

∠BAQ + ∠ABQ + ∠AQB = 180°

⇒ 70° + 70° + ∠AQB = 180°

⇒ ∠AQB = 180° – 140° = 40°.

23. The common point of the tangent and the circle is called _____ .

Answer:

Explanation: point of contact

24. Two concentric circles are of radii 13 cm and 5 cm. The length of the chord of larger circle which touches the smaller circle is _____ .

Answer:

Explanation: 24 cm.

Hint: ∵ AB touches the smaller circle

∴ OC ⊥ AB and hence AC = BC

In right ∆OCA,

OA² = OC² + AC²

⇒ AC² = 13² – 5²

⇒ AC = 12

∴ AB = 2 × 12 = 24 cm

25. A quadrilateral ABCD is drawn to circumscribe a circle. If AB =12 cm, BC = 15 cm and CD = 14 cm, then AD is equal to _____ .

Answer:

Explanation: 11 cm.

Hint: AB + CD = BC + AD

⇒ 12+ 14= 15+ AD

⇒ AD =11 cm.

26. Number of tangents to a circle which are parallel to a secant is ____ .

Answer:

Explanation: 2

27. A tangent PQ at a point P of a circle of radius 7 cm meets a line through centre O at a point Q so that OQ = 25 cm length PQ is ____ .

Answer: 24 cm

Explanation:

24 cm.

Hint: PQ² = OQ² – OP² = 252²- 7²

⇒ PQ = 24 cm.

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