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Question

The circle C1:x2+y2=8 cuts orthogonally the circle C2 whose centre lies on the line xy4=0 then, the circle C2 passes through a fixed point, which lies on

A
x2y=0
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B
x+y=0
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C
x2y=0
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D
x+2y=0
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Solution

The correct option is B x+y=0
The centre of C2 be (g,f).
gf4=0g=f+4
Equation of C2:x2+y22gx2fy+k=0
C2:x2+y22(f+4)x2fy+k=0
C1:x2+y28=0 are orthogonal.
Orthogonality condition -
2×0×(f+4)+2×0×(f)=k8
k=8
C2:x2+y22fx8x2fy+8=0x2+y28x+82f(x+y)=0
Combination of circle x2+y28x+8=0 and line x+y=0.
C2 passes through the point of intersection of the circle and the straight line x+y=0.

y=x & x2+y28x+82f(x+y)=0x2+x28x+8=0(x2)2=0
x=2;y=2 is the point of intersection.

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