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: , 4
Let $S$ be the circle pasing through point of intersection of $\mathrm{S}_{1} \& \mathrm{~S}_{2}$
$\therefore \quad \mathrm{S}=\mathrm{S}_{1}+\lambda \mathrm{S}_{2}=0$
$\Rightarrow S:\left(x^{2}+y^{2}-6 x\right)+\lambda\left(x^{2}+y^{2}-4 y\right)=0$
$\Rightarrow S: x^{2}+y^{2}-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0$
Centre $\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)$ lies on
$2 x-3 y+12=0 \Rightarrow \lambda=-3$
put in $(1) \Rightarrow S: x^{2}+y^{2}+3 x-6 y=0$
Now check options point $(-3,6)$
lies on $\mathrm{S}$.