Question

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

Answer 1

Answer: 12

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
$x^2+y^2-4x-4y+\frac{38}{5}=0$
Now radius of $c_2$ is $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}=r$ ${\left(\text{ radius of }{c}_1\right)}^2={\left(\text{ radius of }{c}_2\right)}^2$
$\Rightarrow 10-\alpha =\frac{2}{5}\Rightarrow \alpha =\frac{48}{5}$
$\therefore \alpha +6r^2=\frac{48}{5}+\frac{12}{5}=12$