Circles Class 10 MCQ with Solutions
1.  In the given figure :
(A) PQ = PR
(B) QR = AB
(C) BQ = BR
(D) AQ = AR
Ans.    (A)
Sol.     PQ = PR (Tangent from the same point on a circle)
2.  Two circles touch each other externally at C and AB is a common tangent to the circles. Then, \angle ACB =
(A) 60º
(B) 45º
(C) 30º
(D) 90º
Ans.    (D)
Sol.    α + θ + α + θ= 180°
2 (α + θ) = 180°
α + θ = 90°
∴  \angle ACB = 90°
3. Two concentric circles are of radii 5 cm and 3 cm. Then the length of the chord of the larger circle which touches the smaller circle is :
(A) 4 cm
(B) 8 cm
(C) 5 cm
(D) 3 cm
Sol.     (B)
In Δ ACO \angle ACO = 90°
OA = 5 cm
OC = 3 cm
AC = \sqrt {A{O^2}\, - \,O{C^2}}
= \sqrt {{5^2}\, - \,{3^2}} \, = \,4\,{\rm{cm}}
AB = 2 × AC = 8 cm
4. How many circles can be drawn passing through three collinear points :
(A) Only one
(B) Two
(C) Infinite
(D) No Circle
Ans.    (D)
Sol.     No circle can be drawn.
5.  In figure, O is centre then \angle AOC is :
(A) 100°
(B) 115°
(C) 130°
(D) 140°
Ans.    (C)
Sol. \angle AOC = (180° – 115°) × 2 = 65° × 2 = 130°
6.  In the figure, O is the centre of circle, BC = OB, \angle ACD = y and \angle AOD = x. Then :
(A) x = y
(B) x = 3y
(C) x = 4y
(D) x = 2y
Ans.    (B)
Sol.     
OB = BC
OAÂ = OB (Radii of same circle) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Let  \angle OAB = \angle ABO = z
∴   z = 2y                        Â
180 – 2z + x + y = 180
x + y = 2z
x + y = 4y
⇒  x = 3y
7.  A circle is inscribed in a quadrilateral PQRS. If QR = 30 cm, QA = 22 cm, RS = 30 cm and PS ⊥ SR then radius of circle is :
(A) 20 cm
(B) 22 cm
(C) 24 cm
(D) 25 cm
Ans.    (B)
Sol.   
Â
So required radius = 22 cm
8.  The length of the tangent from a point A at a circle of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is :
(A) \sqrt 7 cm
(B) 7 cm
(C) 5 cm
(D) 25 cm
Ans.    (C)
Sol.    
OA = \sqrt {O{B^2} + \,A{B^2}} \,\,
= \sqrt {{3^2} + \,{4^2}} \,\,
= \sqrt {25} \,\,
= 5Â cm
9. In the given circle with centre O, PT = 12 cm, PA = 3 cm, PC = 4 cm, find PB + PD.
(A) 80 cm.
(B) 84 cm.
(C) 90 cm.
(D) None
Ans.    (B)
Sol.     PA · PB = PC · PD = PT2
∴  PA · PB = 144
3 · PB = 144
⇒  PB = 48 cm
PC · PD = 144
⇒  PD = 36 cm
∴  PB + PD = 84 cm.
10. What is the maximum number of common tangents that can be drawn to two circles which touches externally?
(A) 0
(B) 1
(C) 2
(D) 3
Ans.    (D)
Sol.    
 Â
11.  In the fig. given below O is the centre of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has a radius of 2 cm. then AC is:
(A) 4 cm
(B) 2 +\sqrt 2 cm
(C) 4 +Â \sqrt 2 cm
(D)Â 2 +2Â \sqrt 2 cm
Ans.    (D)
Sol.     Radius of circle = 2 cm
Join O to B point.
OB is Perpendicular to AB.
\angle OAB = 45° \angle OBA = 90°So, \angle AOB = 45°
AB = OB = 2
ΔOBA is right angled triangle
OA = 2\sqrt 2
AC = OC + OA = (2 + 2\sqrt 2 )\,{\rm{cm}}
12. PQ, PR are tangents to a circle and QS is a diameter, then
(A) \angle QPR = \frac{1}{2}\angle RQS
(B) \angle QPR = 2\angle RQS
(C) \angle QPR =Â \angle RQS
(D) None of these
Ans.    (B)
Sol.     Lets,  O is centre of circle
\angle QPR = α
13.  In the following figure, QS is the diameter and APT the tangent at P. Then \angle APQ is equal to :
(A) 60°
(B) 30°
(C) 40°
(D) 50°
Ans.    (A)
Sol.     In ΔPOQ, OP = OQ (Radius of circle)
\angle OQP = \angle OPQ = 30° \angle APO = 90° \angle APQ = \angle APO – \angle OPQ= 90° – 30° = 60°
14.  In the adjoining figure AOB is a diameter, MPQ is a tangent at P, then the value of \angle MPA is equal to :
(A) 25°
(B) 26°
(C) 27°
(D) 30°
Ans.    (D)
Sol.     \angle AOP + \angle BOP = 180°
\angle AOP = 180° – 120° = 60° \angle OAP = \angle OPA                     (Isosceles Triangle)In ΔAOP
\angle OAP + \angle OPA + \angle AOP = 180°2 \angle OAP = 180° – 60° = 120°
\angle OAP = 60° \angle MPA = \angle MPO – \angle APO= 90° – 60° = 30°
15.  PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then, ratio of area of the square to the area of circle is :
(A) \frac{\pi }{3}
(B) \frac{{11}}{7}
(C) \frac{3}{\pi }
(D) \frac{7}{{11}}
Ans.    (C)
Sol.     Let the radius of circle be r
SO = OT = TR
SR = \sqrt {R{O^2}-O{S^2}}
= \sqrt {4{r^2}-{r^2}}Â = \sqrt 3 \,r \frac{{{\rm{Area of square}}}}{{{\rm{Area of circle}}}}Â = \frac{{\sqrt {\rm{3}} r \cdot \sqrt 3 r}}{{\pi {r^2}}} = \left( {\frac{3}{\pi }} \right)
16.  Two circles of radii a and b touch each other externally. If ST is their common tangent at S and T, then the value of ST2 is equals to :
(A) a – b
(B) (a – b)2
(C) 2ab
(D) 4ab
Ans.    (D)
Sol.     ST2 = (a + b)2 + (a – b)2
= 4ab
17.  In the given figure, circle AXB passes through ‘O’ the centre of circle AYB. AX, BX and AY, BY are tangents to the circles AYB and AXB respectively. The value of y° is :
(A) 180°–x°
(B) 180°– 2x°
(C) \frac{1}{2} (90°–x°)
(D) 90° – \left( {\frac{{x^\circ }}{2}} \right)
Ans.    (D)
Sol.     \angle AOB = 2y°
x° + 2y° = 180°
y° =  90^\circ -\frac{{x^\circ }}{2}
18.  In the figure given below, \angle QSR is equal to :
(A) 65º
(B) 50º
(C) 70º
(D) 75º
Ans.    (A)
Sol.     \angle QOR = 360° – (90° + 90° + 50°) = 130°
\angle QSR = \frac{1}{2}\angle QOR = \frac{1}{2} \times 130^\circ = 65^\circ19. In the adjoining figure (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If \angle ATC = 30° and \angle ACT = 50°, then find \angle BOA .
(A) 60°
(B) 100°
(C) 50°
(D) 120°
Ans.    (B)
Sol.     \angle CAB = 80°
So \angle BOC = 160° and \angle OBC = \angle OCB = 10°
\angle OCT = 90°  ⇒  \angle OCA = 40° \angle BCA = \angle BCO + \angle OCA = 10° + 40° = 50°and \angle BOA = 2 \angle BCA = 100°
20.  A circle is inscribed in a \Delta ABC having sides 8 cm, 10 cm and 12 cm as shown in figure. Find AD.
(A) 5 cm
(B) 4 cm
(C) 6 cm
(D) 7 cm
Ans.    (D)
Sol.     AD = AF = x,
BD = BE = y,  CF = CE = Z
x + y = 12,
y + z = 8, Z + x = 10
x + y + z = 15
on solving x = 7.
21.  In the figure, if O is the centre of the circle, then \angle DOB equals :
(A) 120º
(B) 30º
(C) 90º
(D) 60°
Ans.    (D)
Sol.     \angle ADC = 90°
\angle OAB = 180° – (90° + 60°) = 180° – 150° = 30° \angle OBA = \angle OAB = 30° \angle DOB = \angle OBA + \angle OAB = 30° + 30° = 60°22.  The length of the tangent from a point A at a circle of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is :
(A) \sqrt 7 cm
(B) 7 cm
(C) 5 cm
(D) 25 cm
Ans.    (C)
Sol.     AO2 = AT2 + TO2
AO2Â = 42Â + 32Â = 25
AO = 5 cm
23.  A tangent intersect the circle in :
(A) two points
(B) only at one point
(C) three points
(D) None of these
Ans.    (B)
Sol.     Only at one point.
24.  The incircle of Δ ABC, touches the sides BC, CA and AB at D, E and F respectively then AE + BF + CD =
(A) AB + BC + CA
(B) 2 (AB + BC + CA)
(C)1/2 (AB + BC + CA)
(D) None of these
Ans.    (C)
Sol.     Here AE = AF = x,
BF = BD = y
CE = CD = z
AE + AF + BF + BD + CE + CD = 2x + 2y + 2z
AB + BC + CA = 2(AE + BF + CD)
AE + BF + CD = \frac{1}{2}(AB + BC + CA)
25. In the given figure find \angle ABQ :
(A) 45°
(B) 50°
(C) 65°
(D) 75°
Ans.    (B)
Sol.     \angle OBQ = 90°, \angle OBA = 40°
\angle ABQ = 90° – 40° = 50°26.  In the given figure, TAS is a tangent of the circle, with centre O, at the point A. If \angle OBA = 32°, find the values of x and y :
(A) x = 56°, y = 58°
(B) x =58°, y = 58°
(C) x = 56°, y = 56°
(D) x = 58°, y = 56°
Ans.    (B)
Sol.     \angle OAS = 90°,  \angle OAB = 32°
x = 90° – 32° = 58°
\angle AOB = 180° – 2 × 32° = 180° – 64° = 116°y = \frac{1}{2}\angle AOB = \frac{1}{2} \times 116^\circ = 58^\circ
27.  In the given figure, TBP and TCQ are tangents to the circle whose center is O. Also \angle PBA = 60° and \angle ACQ = 70°. Determine \angle BAC and \angle BTC.
(A) 50°, 60°
(B) 60°,70°
(C) 50°, 70°
(D) 50°, 80°
Ans.    (D)
Sol.     \angle OCQ = \angle OBP = 90°
\angle OCA = \angle OAC = 90° – 70° = 20° \angle OBA = \angle BAO = 90° – 60° = 30° \angle BAC = \angle OAC + \angle BAO \angle BAC = 20° + 30° = 50° \angle BOC = 2 × \angle BAC = 100° \angle BTC = 180° – 100° = 80°28.  In the given figure, quadrilateral ABCD is circumscribed touching the circle at P, Q, R, and S such that \angle DAB = 90°. If CS = 27 cm and CB = 38 cm and the radius of the circle is 10 cm, then find AB.
(A) 22 cm
(B) 10 cm
(C) 21 cm
(D) 20 cm
Ans.    (C)
Sol.     AQOP is a square
AQ = 10 cm
CR = CS = 27 cm
BR = BC – CR = 38 – 27
BR = 11 cm
BQ = BR = 11 cm
AB = AQ + BQ = 10 + 11 = 21 cm
29.  In the figure, if O is the centre of the circle, then \angle BTC equals :
(A) 44°
(B) 113°
(C) 23°
(D) None of these
Ans.    (A)
Sol.     Join OC, \angle OCS = 90°,
\angle OCA = 90° – 67° \angle OCA = 23° \angle OCA = \angle OAC = 23° \angle ACS = \angle CAT + \angle BTC67° = 23° + \angle BTC
\angle BTC = 67° – 23° \angle BTC = 44°30.  The length l  of a tangent drawn from a point A to a circle is \frac{4}{3} of the radius r. The shortest distance from A to the circle is :
(A)Â \frac{1}{2}r
(B) r
(C) \frac{3}{4} r
(D) \frac{2}{3}r
Ans.    (D)
Sol.    
AO2Â = AT2Â + TO2
AO2Â = {\left( {\frac{4}{3}r} \right)^2} + {r^2}
AO2 = \frac{{16{r^2} + 9{r^2}}}{9}\, \Rightarrow \,\frac{{25{r^2}}}{9}
AO = \frac{{5r}}{3}
Shortest distance
AP = AO-OP = \frac{{5r}}{3}-r
AP =Â \frac{{2r}}{3}
31.  In figure, O is centre and \angle ABD = 52º then \angle AOC = :
(A) 72º
(B) 76º
(C) 78º
(D) 82º
Ans.    (B)
Sol.     \angle ADB = 180° – (90° + 52°) = 180° – 142° = 38°
\angle AOC = 2 × \angle ADB = 2 × 38° = 76°32.  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 \angle POA is equal to :
(A) 50°
(B) 60°
(C) 65°
(D) 80°
Ans.    (A)                                                                                                                Â
Sol.     \angle PAO = 90°, \angle APO = 40°
33.  Number of tangents that can be drawn from a point lying inside a circle are :
(A) 0
(B) 1
(C) 2
(D) More than 2
Ans.    (A)
Sol.     No tangent can be drawn
34.  Two perpendicular tangents are drawn from a point P to the circle then figure formed by these tangents and radius drawn on point of contact is :
(A) Rectangle
(B) Square
(C) Rhombus
(D) Trapezium
Ans.    (B)
Sol.     PAOB will be rectangle or square
But AO = BO
i.e. PAOB is a square.
35. The tangents at the end points of a diameter of a circle :
(A) Intersect at a point
(B) Are parallel to each other
(C) Are perpendicular to each other
(D) Meeting at an angle of 45°
Ans.    (B)
Sol.    
 Â
36. In the given figure, ABC is a right angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle :
(A) 2 cm
(B) 1 cm
(C) 1.5 cm
(D) 2.5 cm
Ans.    (A)
Sol.     AL = AN = r
LB = MB = 6 – r
CN = CM = 8 – r
BC = CM + MB = 6 – r + 8 – r
BC = 14 – 2r
BC = 10
14 – 2r = 10
r = \frac{{14-10}}{2} = 2\,cm
37. If ΔABC is circumscribed touching the circle at P, Q and R. If AP = 4 cm, BP = 6 cm and AC= 12 cm, and
BC = x cm. Find the value of x.
(A) 14 cm
(B) 15 cm
(C) 16 cm
(D) 13 cm.
Ans.    (A)
Sol.     AP = AR = 4 cm
BP = BQ = 6 cm
CR = AC – AR = 12 – 4 = 8 cm
BC = BQ + CQ
= 6 + CR = 6 + 8 = 14 cm
38.  If three circles with centers A, B, C respectively touch each other externally. If AB = 5cm, BC = 7cm and CA = 6 cm. Then find the radius of the circle with centre A.
(A) 3 cm
(B) 2 cm
(C) 2.5 cm
(D) 3.5
Ans.    (B)
Sol.     AP = AR = x, BP = BQ = y, CQ = CR = z
AB = x + y = 5     …..(1)
BC = y + z = 7    …..(2)
CA = z + x = 6     …..(3)
adding (1) and (3)
2x + y + z = 11
from (2)
2x + 7 = 11
2x = 4
x = 2
radius of the circle with centre A = 2 cm.
39. In the adjoining figure, AOB is a chord of a circle ABC, T is a point on the tangent at A, the tangents at B meets AT produced to D. \angle ATO = 55°, \angle BOT = 120°, find \angle BPT:
(A) 5°
(B) 10°
(C) 12°
(D) 15°
Ans.    (A)
Sol.     Let O’ is a centre of a circle
\angle OAT = 120° – 55° = 65° \angle O’AT = 90°
40.  PT is a tangent to a circle drawn from an external point P and PBA is a secant. If AP = 10 cm, BP = 8 cm then PT =
(A) 3 cm
(B) 4 cm
(C) 5 cm
(D) 4\sqrt 5 cm
Ans.    (D)
Sol.     Here PT2 = PA × PB
PT2 = 10 × 8
PT2 = 80
PT = \sqrt {16 \times 5}Â = 4\sqrt 5 {\rm{ cm}}
