Question

Let the mirror image of a circle
$c_1: x^2+y^2-2 x-6 y+\alpha=0 \text { in line } y=x+1$
be $c_2: 5 x^2+5 y^2+10 g x+10 f y+38=0$.
If $r$ is the radius of circle $c_2$, then $\alpha+6 r ^2$ is equal to

Answer 1

Image of centre $c_1 \equiv(1,3)$ in $x-y+1=0$ is given by
$ \frac{x_1-1}{1}=\frac{y_1-3}{-1}=\frac{-2(1-3+1)}{1^2+1^2} $
$\Rightarrow x_1=2, y_1=2$
$\therefore$ Centre of circle $c _2 \equiv(2,2)$
$\therefore$ Equation of $c _2$ be
$x^2+y^2-4 x-4 y+\frac{38}{5}=0$
Now radius of $c_2$ is $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}=r$ $\left(\text { radius of } c_1\right)^2=\left(\text { radius of } c_2\right)^2$
$ \Rightarrow 10-\alpha=\frac{2}{5} \Rightarrow \alpha=\frac{48}{5} $
$ \therefore \alpha+6 r ^2=\frac{48}{5}+\frac{12}{5}=12$