The circle passing through the intersection o | vedclass

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Question

Let $S$ be the circle pasing through point of intersection of $S _{1} \& S _{2}$ $	herefore S = S _{1}+\lambda S _{2}=0$

$\Rightarrow S :\left

Vedclass · Accepted Answer

$(-3,6)$

Vedclass · Answer

Let $S$ be the circle pasing through point of intersection of $S _{1} \& S _{2}$ $	herefore S = S _{1}+\lambda S _{2}=0$

$\Rightarrow S :\left( x ^{2}+ y ^{2}-6 x ight)+\lambda\left( x ^{2}+ y ^{2}-4 y ight)=0$

$\Rightarrow S : x ^{2}+ y ^{2}-\left(\frac{6}{1+\lambda}ight) x -\left(\frac{4 \lambda}{1+\lambda}ight) y =0 \ldots(1)$

Centre $\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}ight)$ 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 $S$.

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

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