Question

# If the circle x2+y2+2gx+2fy+c=0 cuts each of the circles x2+y2=4;x2+y2âˆ’6xâˆ’8y+10=0 andx2+y2+2xâˆ’4yâˆ’2=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=0s1:x2+y2=4 centre=(0,0)s2:x2+y2=6x−8t+10=0 centre =(3,4)s3:x2+y2+2x−4y−2=0 centre=(−1,2)So, radial axis of s and s1 passes through (0,0)2gx+2fy+c+4=0c=−4So, radical axis of s and s2 passes through (Q,1).(2g+6)x+(28+8)y−14=06g+18+8f+18=03g+4f+18=0−−(1)So, radical axis of s and s3 pass through (−1,2)(29−2)x+(2f+4)y−2=0−2g+2+4f+6=02f−g+4=0−−(1)from (1) & (2)10f+30=0f=−3 and g=−2So, centre of circle is (−g,−f)=(2,3)

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