Let the mirror image of a circle ${\mathrm{c}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-6\mathrm{y}+\mathrm{\alpha }=0$ in line $\mathrm{y}=\mathrm{x}+1\text{ be }{\mathrm{c}}_2:5{\mathrm{x}}^2+5{\mathrm{y}}^2+10\mathrm{gx}+10\mathrm{fy}+38=0$. If $\mathrm{r}$ is the radius of circle c₂, then $\mathrm{\alpha }+6{\mathrm{r}}^2$is equal to_______.

${\mathrm{c}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-6\mathrm{y}+\mathrm{\alpha }=0$
$\text{ Centre of }{\mathrm{c}}_1\text{ i.e., }\mathrm{A}\equiv (1,3)\text{ and radius }=\sqrt{1+9-\mathrm{\alpha }}=\sqrt{10-\mathrm{\alpha }}$
$\begin{array}{l}{\mathrm{c}}_2:5{\mathrm{x}}^2+5{\mathrm{y}}^2+10\mathrm{gx}+10\mathrm{fy}+38=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\frac{38}{5}=0\end{array}$
${\text{ Centre of c}}_2 \mathrm{i}.\mathrm{e}., \mathrm{B}\equiv (-\mathrm{g},-\mathrm{f}) \mathrm{and} \mathrm{radius} =\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\frac{38}{5}}$
Image of centre A in x - y+ 1= 0 is given by $\frac{-\mathrm{g}-1}{1}=\frac{-\mathrm{f}-3}{-1}=\frac{-2(1-3+1)}{1^2+1^2}$
$\begin{array}{l}\Rightarrow -\mathrm{g}-1=\mathrm{f}+3=1\Rightarrow \mathrm{g}=-2,\mathrm{f}=-2\\ \therefore \mathrm{r}=\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}\Rightarrow \boxed{{\mathrm{r}}^2=\frac{2}{5}}\end{array}$

$\text{ Also, radius of }{\mathrm{c}}_1=\text{ radius of }{\mathrm{c}}_2$
$\Rightarrow \sqrt{10-\mathrm{\alpha }}=\sqrt{\frac{2}{5}}\Rightarrow 10-\mathrm{\alpha }=\frac{2}{5}\Rightarrow \boxed{\mathrm{\alpha }=\frac{48}{5}}$
Thus, $$\begin{gathered} \mathrm{\alpha }+6{\mathrm{r}}^2=\frac{48}{5}+6\left(\frac{2}{5}\right)=\frac{60}{5}=12 \\ \therefore \boxed{\mathrm{\alpha }+6{\mathrm{r}}^2=12} \end{gathered}$$