# 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

Explanation:

Reason: Here radius, r = 4 cm
Required distance,
AB = OA + OB
= r + r = 2r = 2Ã—4 = 8 cm

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

(a) 135Â°
(b) 145Â°
(c) 165Â°
(d) 155Â°

Explanation: Reason: Since OR âŠ¥ PR and OS âŠ¥ PS
âˆ´ âˆ ORP = âˆ OSP = 90Â°
In â–¡ ORPS, âˆ ROS + âˆ ORP + âˆ RPS + âˆ OSP = 360Â°
âˆ ROS + 90Â° + 25Â° + 90Â° = 360Â°
âˆ ROS = 360Â° â€“ 205Â° = 155Â°

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

Explanation:

Reason: In right angle âˆ†OAP, APÂ² + OAÂ² = OPÂ²
â‡’ APÂ² + (8)Â² = (17)Â² => APÂ² + 64 = 289
â‡’ APÂ² = 289 â€“ 64 = 225
âˆ´ AP = âˆš225 = 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

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

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

Explanation:

Reason: Since OP âŠ¥ AB and OQ âŠ¥ CD
âˆ´ Z1 = 90Â° and Z2 = 90Â°
â‡’ âˆ 1 = Z2, which are alternate angles.
âˆ´ AB || CD

7. Tangents from an external point to a circle are
(a) equal
(b) not equal
(c) parallel
(d) perpendicular

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

Explanation:

Reason: In right angle Î”OAP, we have
OAÂ² + APÂ² = OPÂ²
â‡’ OAÂ² + (8)Â² = (10)2
â‡’ OAÂ² + 64 = 100
â‡’ OAÂ² = 100 â€“ 64 = 36
âˆ´ OA = âˆš36 = 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

Explanation: Reason: Since
BQ = BR â€¦(i) [âˆµ Tangents drawn from external points are equal]
CQ = CP â€¦[Using (i)]
BC + BQ = 11
â‡’ 6 + BR = 11
â‡’ BR = 11 â€“ 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 cmÂ²
(b) 65 cmÂ²
(c) 30 cmÂ²
(d) 32.5 cmÂ²

Explanation:

Reason: OPÂ² = OQÂ² + PQÂ²
169 = 25 + PQÂ²
PQÂ² = 144
PQ = 12
Area PQOR = ar (AOPQ) + ar (AOPR)
= 1/2 Ã— 12 Ã— 5 + 1/2 Ã— 12 Ã— 5 = 60 cmÂ²

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

(a) 40Â°
(b) 50Â°
(c) 140Â°
(d) 150Â°

Explanation:
âˆ ABO + âˆ BOC + âˆ OCA + âˆ BAC = 360Â°
â‡’ 90Â° + âˆ BOC + 90Â° + 40Â° = 360Â°
â‡’ âˆ BOC = 360Â° â€“ 220Â° = 140Â°

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

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

âˆ´ OT âŠ¥ PT
Now, in AOTP,
â‡’ OPÂ² = PTÂ² + OTÂ²
â‡’ 26Â² = 24Â² + OTÂ²
â‡’ 676 â€“ 576 = OTÂ²
â‡’ 100 = OTÂ²
â‡’ 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

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 50Â° with PQ, then âˆ POQ is equal to [NCERT Exemplar Problems]

(a) 100Â°
(b) 80Â°
(c) 90Â°
(d) 75Â°

Explanation:
(a) OP âŠ¥ PR [Y Tangent and radius are âŠ¥ to each other at the point of contact]
âˆ OPQ = 90Â° â€“ 50Â° = 40Â°
âˆ´ âˆ OPQ = âˆ OQP = 40Â°
In âˆ†OPQ,
â‡’ âˆ POQ + âˆ OPQ + âˆ OQP = 180Â°
â‡’ âˆ POQ + 40Â° + 40Â° = 180Â°
âˆ POQ = 180Â° â€“ 80Â° = 100Â°.

15. In figure AT is a tangent to the circle with centre O such that OT = 4 cm and âˆ OTA = 30Â°. Then AT is equal to [NCERT Exemplar Problems]
(a) 4 cm
(b) 2 cm
(c) 2âˆš3 cm
(d) 4âˆš3 cm

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

Explanation:
(c) Let radius of circle = R
âˆ´ Ï€RÂ² = 25Ï€
â‡’ R = 5 cm
âˆ´ Distance between two parallel tangents = diameter = 2 Ã— 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

Explanation:

(c) âˆµ AB touches
C(0, r<sub>2</sub>)
âˆ´ OC âŠ¥ AB

Also, perpendicular from the centre to a chord bisects the chord.
âˆ´ AC = BC

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

(a) 100Â°
(b) 40Â°
(c) 50Â°
(d) 90Â°

Explanation:
(c) âˆ AOB = 100Â°
âˆ OAB = âˆ OBA (âˆµ OA and OB are radii)
Now, in âˆ†AOB,
âˆ AOB + âˆ OAB + âˆ OBA = 180Â°
(Angle sum property of A)
â‡’ 100Â° + x + x = 180Â° [Let âˆ OAB = âˆ OBA = x]
â‡’ 2x = 180Â° â€“ 100Â°
â‡’ 2x = 80Â°
â‡’ x = 40Â°
Also, âˆ OAB + âˆ BAT = 90Â°
[âˆµ OA is radius and TA is tangent at A]
â‡’ 40Â° + âˆ  BAT = 90Â°
â‡’ âˆ BAT = 50Â°

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

(a) 30Â°
(b) 60Â°
(c) 90Â°
(d) 15Â°

Explanation:
(a) Given âˆ APB = 60Â°
âˆµ âˆ APB + âˆ PAB + âˆ PBA = 180Â°
â‡’ APB + x + x = 180Â°
[âˆµ PA = PB âˆ´ âˆ PAB = âˆ PBA = x (say)]
â‡’ 60Â° + 2x = 180Â°
â‡’ 2x = 180Â° â€“ 60Â°
â‡’ 2x = 120Â°
â‡’ x = 120Â°/2 = 60Â°
Also, âˆ OAP = 90Â°
â‡’ âˆ OAB + âˆ PAB = 90Â°
â‡’ âˆ OAB + 60Â°= 90Â°
â‡’ âˆ OAB = 30Â°

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

(a) 55Â°
(b) 70Â°
(c) 110Â°
(d) 90Â°

Explanation:
âˆ PTQ + âˆ TPO + âˆ TQO + âˆ POQ
= 360Â°
â‡’ âˆ PTQ + 90Â° + 90Â° + 110Â° = 360Â°
â‡’ âˆ PTQ + 290Â° = 360Â°
â‡’ âˆ PTQ = 360Â° â€“ 290Â° = 70Â°

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

(a) 63Â°
(b) 153Â°
(c) 110Â°
(d) 90Â°

Explanation:

(c) âˆ QPA = âˆ RPA
[âˆµ âˆ†AQP â‰… âˆ†ARP (RHS congruence rule)]
â‡’ âˆ RPA = 27Â°
âˆ´ âˆ QPR = âˆ QPA + âˆ RPA
= 27Â° + 27Â° = 54Â°
Now,
âˆ QAR + âˆ AQP + âˆ ARP + âˆ QPR = 360Â°
â‡’ âˆ QAR = 90Â° + 90Â° + 54Â° = 360Â°
â‡’ âˆ QAR = 360Â° â€“ 234Â° = 126Â°

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 âˆ BQR = 70Â°, then âˆ AQB is equal to [NCERT Exemplar Problems]

(a) 20Â°
(b) 40Â°
(c) 35Â°
(d) 45Â°

Explanation:
(b) AB || PR
âˆ´ âˆ ABQ =âˆ BQR
[Alternate interior angles]
â‡’ âˆ ABQ = 70Â°
Also, âˆ BQR = âˆ BAQ [Angles in alternate segment]
â‡’ âˆ BAQ = 70Â°
In âˆ†AQB,
âˆ BAQ + âˆ ABQ + âˆ AQB = 180Â°
â‡’ 70Â° + 70Â° + âˆ AQB = 180Â°
â‡’ âˆ AQB = 180Â° â€“ 140Â° = 40Â°.

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

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 _____ .

Explanation: 24 cm.

Hint: âˆµ AB touches the smaller circle

âˆ´ OC âŠ¥ AB and hence AC = BC
In right âˆ†OCA,
OAÂ² = OCÂ² + ACÂ²
â‡’ ACÂ² = 13Â² â€“ 5Â²
â‡’ AC = 12
âˆ´ AB = 2 Ã— 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 _____ .

Explanation: 11 cm.

Hint: AB + CD = BC + AD

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

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 ____ .

Explanation:

24 cm.
Hint: PQÂ² = OQÂ² â€“ OPÂ² = 252Â²- 7Â²
â‡’ PQ = 24 cm.

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