If the equation of the circle passing through the points of intersection of the circles

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Aicharya · Accepted Answer

$S_1=x^2-2 x+y^2-4 y-4=0$ $S_2=x^2+2 x+y^2+4 y-4=0$ Equation of circle passing through intersection of $\mathrm{S}_1$ and $\mathrm{S}_2$ is $S_1+\lambda S_2=0$ $\begin{aligned} \Rightarrow(1+\lambda) x^2+2(\lambda-1) x+(1+\lambda) y^2+4(\lambda-1) \ y-4-4 \lambda=0\end{aligned}$ Also it passes $(3,3)$ $\Rightarrow 9(1+\lambda)+6(\lambda-1)+9(1+\lambda)+12(\lambda-1)-4-4 \lambda=0$ $\Rightarrow 18(1+\lambda)+18(\lambda-1)-4-4 \lambda=0 \Rightarrow \lambda=\frac{1}{8}$ Now, $\alpha=\frac{2(\lambda-1)}{(1+\lambda)}=\frac{-14}{9} \Rightarrow \beta=\frac{4(\lambda-1)}{(1+\lambda)}=\frac{-28}{9}$ $\gamma=\frac{-4(\lambda+1)}{(1+\lambda)}=-4 ;$ Now, $3(\alpha+\beta+\gamma)=3\left(\frac{-78}{9}ight)=-26$

If the equation of the circle passing through the points of intersection of the circles

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