Class 10 Maths MCQs Chapter 10 Circles

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, r₁) and C(O, r₂) are two concentric circles with r₁ > r₂ AB is a chord of C(O, r₁) touching C(O, r,₂) at C then
(a) AB = r₁
(b) AB = r₂
(c) AC = BC
(d) AB = r₁ + r₂

Answer: c
Explanation:

(c) ∵ AB touches
C(0, r₂)
∴ 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.