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

<p>${c_1}:{x^2} + {y^2} - 2x - 6y + \alpha = 0$</p> <p>Then centre $= (1,3)$ and radius $(r) = \sqrt {10 - \alpha }$</p> <p>Image of $(1,3)$ w.r.t. line $x - y + 1 = 0$ is $(2,2)$</p> <p>${c_2}:5{x^2} + 5{y^2} + 10gx + 10fy + 38 = 0$</p> <p>or ${x^2} + {y^2} + 2gx + 2fy + {{38} \over 5} = 0$</p> <p>Then $( - g, - f) = (2,2)$</p> <p>$\therefore$ $g = f = - 2$ .......... (i)</p> <p>Radius of ${c_2} = r = \sqrt {4 + 4 - {{38} \over 5}} = \sqrt {10 - \alpha }$</p> <p>$\Rightarrow {2 \over 5} = 10 - \alpha$</p> <p>$\therefore$ $\alpha = {{48} \over 5}$ and $r = \sqrt {{2 \over 5}}$</p> <p>$\therefore$ $\alpha + 6{r^2} = {{48} \over 5} + {{12} \over 5} = 12$</p>