The circle passing through the intersection of the circles,
Question:

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:

1. (1) $(-1,3)$

2. (2) $(-3,6)$

3. (3) $(-3,1)$

4. (4) $(1,-3)$

Correct Option: , 2

Solution:

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).