Mathematics - Circle Question with Solution | TestHub

Let the mirror image of a circle $c_1:x^2+y^2-2x-6y+\alpha =0$  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 ______

Now Image of centre $c_1\equiv \left(1,3\right)$ in $x-y+1=0$ is given by

$\frac{x_1-1}{1}=\frac{y_1-3}{-1}=\frac{-2\left(1-3+1\right)}{1^2+1^2}$

$\Rightarrow x_1=2,y_1=2$

So, centre of circle $c_2\equiv \left(2,2\right)$

Now radius of circle $c_2$ will be  $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}=r$

Now $\left(\text{radius of} {c_1}^2\right)=$${\left(\text{radius of} c_2\right)}^2$ as they are image to one another,

$\Rightarrow \sqrt{1^2+3^2-\alpha }=\sqrt{\frac{2}{5}}$

$\Rightarrow 10-\alpha =\frac{2}{5}\Rightarrow \alpha =\frac{48}{5}$

Now putting the value of $\alpha \& r$ in $\alpha +6r^2$ we get,

$\alpha +6r^2=\frac{48}{5}+\frac{12}{5}=12$