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Question

If the circle x2+y2+2gx+2fy+c=0 cuts each of the circles x2+y2=4;
x2+y26x8y+10=0 and
x2+y2+2x4y2=0 at the extremities of a diameter, then its centre is

A
(2,3)
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B
(2,3)
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C
(2,3)
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D
(2,3)
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Solution

The correct option is A (2,3)
s:x2+y2+2gx+2fy+c=0
s1:x2+y2=4 centre=(0,0)
s2:x2+y2=6x8t+10=0 centre =(3,4)
s3:x2+y2+2x4y2=0 centre=(1,2)
So, radial axis of s and s1 passes through (0,0)
2gx+2fy+c+4=0
c=4
So, radical axis of s and s2 passes through (Q,1).
(2g+6)x+(28+8)y14=0
6g+18+8f+18=0
3g+4f+18=0(1)
So, radical axis of s and s3 pass through (1,2)
(292)x+(2f+4)y2=0
2g+2+4f+6=0
2fg+4=0(1)
from (1) & (2)
10f+30=0
f=3 and g=2
So, centre of circle is (g,f)=(2,3)

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