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Question

Find the equation of the circle passing through the points of intersection of the circles x2+y24x6y12=0 and x2+y2+6x+4y12=0 and intersection the circle x2+y22x4=0 orthogonally.

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Solution

Let, the equation of the circle through the intersection of given circle is,
x2+y24x6y+12=0(1)
Here, the equation of radical axis for the circles
x2+y24x6y+12=0
and x2+y2+6x+4y12=0 is,
10x10y=0
So, equation (1) can be written as,
x2+y24x6y12+λ(10x10y)=0x2+y2x(10λ+4)y(10λ+6)12=0
It cuts the circle v orthogonally.
2gg1+2ff1=c+c1(10λ+4)(1)+(10λ+6)(0)=124λ=2
Hence, the required equation of the circle is,
x2+y24x6y122(10x10y)=0x2+y2+16x+14y12=0.

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