# Class 11 Maths MCQ – Conic Sections

This set of Class 11 Maths Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Conic Sections”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Find the equation of circle with center at origin and radius 5 units.
a) x2+y2=25
b) x2+y2=5
c) x2=25
d) y2=25

Explanation: Equation of circle with center at (a, b) and radius r units is
(x-a)2+(y-b)2=r2
So, equation of circle is (x-0)2+(y-0)2=52 => x2+y2=25.

2. Find the equation of circle with center at (2, 5) and radius 5 units.
a) x2+y2+4x-10y+4=0
b) x2+y2-4x-10y+4=0
c) x2+y2+4x+10y+4=0
d) x2+y2+4x-10y-4=0

Explanation: Equation of circle with center at (a, b) and radius r units is
(x-a)2+(y-b)2=r2
So, equation of circle is (x-2)2+(y-5)2=52 => x2+y2-4x-10y+4=0.

3. Find the center of the circle with equation x2+y2-4x-10y+4=0.
a) (-2, 5)
b) (-2, -5)
c) (2, -5)
d) (2, 5)

Explanation: Comparing the equation with general form x2+y2+2gx+2fy+c=0, we get
2g=-4 => g=-2
2f=-10 => f=-5
c=4
Center is at (-g, -f) i.e. (2, 5).

4. Find the radius of the circle with equation x2+y2-4x-10y+4=0.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Explanation: Comparing the equation with general form x2+y2+2gx+2fy+c=0, we get
2g=-4 => g=-2
2f=-10 => f=-5
c=4
Radius = $$\sqrt{g^2+f^2-c} = \sqrt{4+25-4}$$=5.

5. Find the equation of circle which pass through (5, 9) and center at (2, 5).
a) x2+y2+4x-10y+4=0
b) x2+y2-4x-10y+4=0
c) x2+y2+4x+10y+4=0
d) x2+y2+4x-10y-4=0

Explanation: Equation of circle with center at (a, b) and radius r units is
(x-a)2 + (y-b)2 = r2
(5-2)2 + (9-5)2 = r2 => r2=32+42 => r=5.
So, equation of circle is (x-2)2+(y-5)2=52 => x2+y2-4x-10y+4=0.

6. If a circle pass through (2, 0) and (0, 4) and center at x-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Explanation: Equation of circle with center at x-axis (a, 0) and radius r units is
(x-a)2+(y)2=r2
=>(2-a)2+(0)2=r2
And (0-a)2+(4)2=r2
=>(a-2)2=a2+42 => (-2)(2a-2) =16 => a-1=-4 => a=-3
So, r2 = (2+3)2=52
r=5 units.

7. If a circle pass through (4, 0) and (0, 2) and center at y-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Explanation: Equation of circle with center at y-axis (0, b) and radius r units is
(x)2+(y-b)2=r2
=>(4)2+(-b)2=r2
And (0)2+(2-b)2=r2
=>(b-2)2=b2+42 => (-2)(2b-2)=16 => b-1=-4 => b=-3
So, r2=42+32=52 => r=5 units.

8. The point (1, 4) lie ___________ the circle x2+y2-2x-4y+2=0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Explanation: Circle has equation x2+y2-2x-4y+2=0.
12+42-2*1-4*4+2 = 1+16-2-16+2 =1 > 0 so, point is outside the circle.

9. The point (0, 0) lie ___________ the circle x2+y2-2x-4y=0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Explanation: Circle has equation x2+y2-2x-4y=0.
02+02-2*0-4*0+0 = 0 so, point is on the circle.

10. The point (6, 2) lie ___________ the circle x2+y2-2x-4y-36=0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Explanation: Circle has equation x2+y2-2x-4y-36=0.
62+22-2*6-4*2-36 = 36+4-12-8-36 =-16<0 so, point is inside the circle.

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