The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$, having its centre on the line, $2 x-3 y+12=0$, also passes through the point:
Correct Option: , 2
We know family of circle be $S_{1}+\lambda S_{2}=0$
$x^{2}+y^{2}-6 x+\lambda\left(x^{2}+y^{2}-4 y\right)=0$
$\Rightarrow(1+\lambda) x^{2}+(1+\lambda) y^{2}-6 x-4 \lambda y=0$....(1)
Centre $(-g,-f)=\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{\lambda+1}\right)$
Centre lies on $2 x-3 y+12=0$, then
$\frac{6}{\lambda+1}-\frac{6 \lambda}{\lambda+1}+12=0 \Rightarrow \lambda=-3$
Equation of circle (i),
$-2 x^{2}-2 y^{2}-6 x+12 y=0$
$\Rightarrow x^{2}+y^{2}+3 x-6 y=0$ (2)
Only $(-3,6)$ satisfy equation (ii).
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