A Circle S passes through the points of intersection of the circles $x^2+y^2-2 x+2 y-2=0$ and $x^2+y^2+2 x-2…
A Circle S passes through the points of intersection of the circles $x^2+y^2-2 x+2 y-2=0$ and $x^2+y^2+2 x-2 y+1=0$. If the centre of this circle S lies on the line $x-y+6=0$, then the radius of the circle $S$ is
$\sqrt{5}$
$5$
$\sqrt{41}$
$\sqrt{14}$
Solution
Since equation of circles is $x^2+y^2-2 x+2 y-2+$
$\begin{aligned}& k\left(x^2+y^2+2 x-2 y+1\right)=0\\&\Rightarrow(1+k) x^2+(1+k) y^2+(2 k-2) x+(2-2 k) y+k-2=0\end{aligned}$
$\Rightarrow x^2+y^2+\frac{2(k-1)}{k+1} x+\frac{2(1-k)}{1+k} y+\frac{k-2}{k+1}=0$
So, centre $=\left(\frac{-(k-1)}{k+1}, \frac{-(1-k)}{1+k}\right)=\left(\frac{1-k}{1+k}, \frac{k-1}{k+1}\right)$
Since, centre lies on the line $x-y+6=0$
$\Rightarrow \frac{1-k}{1+k}-\frac{(k-1)}{1+k}+6=0 \Rightarrow \frac{1-k-k+1}{1+k}=-6$
$\begin{aligned} & \Rightarrow 2-2 k=-6(1+k) \\ & \Rightarrow 1-k=-3-3 k \Rightarrow k=-2\end{aligned}$
So, radius $=\sqrt{\left(\frac{1-k}{1+k}\right)^2+\left(\frac{k-1}{k+1}\right)^2-\left(\frac{k-2}{k+1}\right)}$
$=\sqrt{(-3)^2+(3)^2-4}=\sqrt{14}$