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If a circle ' $S$ ' passing through the origin and having its centre on the line $x-y=0$ cuts the circle $x^2+y^2-4 x-6 y$ $+10=0$ orthogonally, then the diameter of 'S' is
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The correct answer is:
$2 \sqrt{2}$
$\because$ Circle passing through origin. $\therefore c=0$
Centre lies on $y=x \Rightarrow$ centre $(-g,-g)$
$\begin{aligned} & x^2+y^2+2 g x+2 g y=0 \\ & \therefore \quad 2 g(-2)+2 g(-3)=10 \Rightarrow-10 g=10 \Rightarrow g=-1 \\ & \therefore \quad \text { Radius }=\sqrt{1^2+1^2}=\sqrt{2} \\ & \text { Diameter }=2 \sqrt{2} .\end{aligned}$
Centre lies on $y=x \Rightarrow$ centre $(-g,-g)$
$\begin{aligned} & x^2+y^2+2 g x+2 g y=0 \\ & \therefore \quad 2 g(-2)+2 g(-3)=10 \Rightarrow-10 g=10 \Rightarrow g=-1 \\ & \therefore \quad \text { Radius }=\sqrt{1^2+1^2}=\sqrt{2} \\ & \text { Diameter }=2 \sqrt{2} .\end{aligned}$
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