# The circle is passing through the intersection of the circle, $x^{2}+y^{2}-6x=0$ and $x^{2}+y^{2}-4y=0$, having its centre on the line, $2x-3y+12=0$, also passes through the point: Option: 1 (-1,3) Option: 2 (-3,6) Option: 3 (-3,1) Option: 4 (1,-3)

Let S be the circle passing through point of intersection of S1 & S2

\begin{aligned} &\therefore \mathrm{S}=\mathrm{S}_{1}+\lambda \mathrm{S}_{2}=0\\ &\Rightarrow \mathrm{S}:\left(\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}\right)+\lambda\left(\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{y}\right)=0\\ &\Rightarrow \mathrm{S}: \mathrm{x}^{2}+\mathrm{y}^{2}-\left(\frac{6}{1+\lambda}\right) \mathrm{x}-\left(\frac{4 \lambda}{1+\lambda}\right) \mathrm{y}=0 \;\;\;\;\;\;\;\;\;\;\;\ldots(1) \end{aligned}

$\\\text{Centre} \left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right) \text{ lies on} \\2 x-3 y+12=0 \Rightarrow \lambda=-3\\$

Put in (1)

$\Rightarrow \mathrm{S}: \mathrm{x}^{2}+\mathrm{y}^{2}+3 \mathrm{x}-6 \mathrm{y}=0$

Now check options point (–3, 6)

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