Mathematics Questions and Answers – Conic Sections – Circle

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Conic Sections – Circle”.

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
View Answer

Answer: a
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.
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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
View Answer

Answer: b
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)
View Answer

Answer: d
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).
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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
View Answer

Answer: c
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
View Answer

Answer: b
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.
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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
View Answer

Answer: c
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
View Answer

Answer: c
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.
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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
View Answer

Answer: b
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
View Answer

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

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

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter