If $S \equiv 2 x^2+2 y^2-8 x+8 y-7=0$ is the circle passing through the points of intersection of the…
If $S \equiv 2 x^2+2 y^2-8 x+8 y-7=0$ is the circle passing through the points of intersection of the circles $x^2+y^2+$ $\mathrm{kx}-\mathrm{ky}+1=0$ and $\mathrm{x}^2+\mathrm{y}^2-\mathrm{kx}+\mathrm{ky}-2=0$, then the length of the tangent drawn from the point $(k, k)$ to the circle $\mathrm{S}$ is
$\frac{3}{\sqrt{2}}$
3
$\sqrt{\frac{23}{2}}$
$\sqrt{23}$
Solution
The equation of circle through points of intersection of $S_1, S_2$ is:
$\begin{aligned} & \mathrm{S}_1+\lambda\left(\mathrm{S}_2-\mathrm{S}_1\right)=0 \\ & \Rightarrow x^2+y^2+k x-k y+1+\lambda\left(x^2+y^2-k x+k y-2-x^2\right. \\ & \left.\quad-y^2-k x+k y-1\right)=0 \\ & \Rightarrow x^2+y^2+k x-k y+1+\lambda(-2 k x+2 k y-3)=0 \\ & \Rightarrow x^2+y^2+(1-2 \lambda) k x-(1-2 \lambda) k y+(1-3 \lambda)=0\end{aligned}$ ...(i)
Given equation of circle through points of intersection of $\mathrm{S}_1 \& \mathrm{~S}_2$ is $2 x^2+2 y^2-8 x+8 y-7=0$
or, $x^2+y^2-4 x+4 y-\frac{7}{2}=0$ ...(ii)
Comparing eq ${ }^{\mathrm{n}}$ (i) \& (ii), we get $k(1-2 \lambda)=-4 \&-k(1-2 \lambda)=4$
$\& 1-3 \lambda=-\frac{7}{2} \Rightarrow 3 \lambda=\frac{9}{2} \Rightarrow \lambda=\frac{3}{2}$
In $\triangle \mathrm{PMC}, \mathrm{PM}^2+\mathrm{CM}^2=\mathrm{CP}^2$
$\begin{aligned} & \Rightarrow \mathrm{PM}^2+\frac{23}{2}=(2-2)^2+(2+2)^2 \\ & \Rightarrow \mathrm{PM}^2=16-\frac{23}{2}=\frac{9}{2} \Rightarrow \mathrm{PM}=\frac{3}{\sqrt{2}}\end{aligned}$
