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  4. Equation of the circle passing through the point of intersection of the circles x 2 +y2=6 and x2+y2-6 x+8=0 and the point (1,1) is

Q. Equation of the circle passing through the point of intersection of the circles $x ^{2}$ $+y^{2}=6$ and $x^{2}+y^{2}-6 x+8=0$ and the point $(1,1)$ is

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Solution:

Equation of the circle passing through the intersection is
$S_{1}+ kS _{2}=0$.
$\Rightarrow x ^{2}+ y ^{2}-6+ k \left( x ^{2}+ y ^{2}-6 x +8\right)=0$ . . . . . . . (1)
Since the required circle passes through $(1,1)$
$1+1-6+k(1+1-6+8)=0 \Rightarrow k=1$.
$\therefore$ From (1) equation is $x^{2}+y^{2}-3 x+1=0$.

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