Consider a circle C 1   :   x   2 + y 2   –   4 x   –   2 y…

Given,

$C_1 : x_2+ y_2–4x– 2y+\left(5 – \alpha \right)= 0$

So, its centre will be, $O_1=\left(2, 1\right)$ and radius $=\sqrt{\alpha }$

And $C_2 : 5x^2+5y^2–10fx–10gy+36=0$

$\Rightarrow C_2 : x^2+y^2–2fx–2gy +\frac{36}{5}=0$

So, Centre $O_2 = \left(f, g\right)$ and radius  $r=\sqrt{f^2+g^2-\frac{36}{5}}$

Also given $O_2$ is reflection of $O_1$ in $2x–y+1=0$, so image formula we get,

$\Rightarrow \frac{f-2}{2}=\frac{g-1}{-1}=-2·\left(\frac{2\times 2-1+1}{2^2+1^2}\right)$

$\Rightarrow f=\frac{-6}{5}$ and $g=\frac{13}{5}$

So, radius $r=\sqrt{{\left(\frac{-6}{5}\right)}^2+{\left(\frac{13}{5}\right)}^2-\frac{36}{5}}=\frac{\sqrt{25}}{5}=1$

$\Rightarrow r=1$and $\alpha =1$ as they both are same radius circle,

Hence, $r+\alpha =2$.