Consider a circle ${\mathrm{C}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-2\mathrm{y}=\mathrm{\alpha }-5$. Let its mirror image in the line y = 2x + 1 be another circle ${\mathrm{C}}_2:5{\mathrm{x}}^2+5{\mathrm{y}}^2-10\mathrm{fx}-10\mathrm{gy}+36=0.$ Let ‘r’ be the radius of C₂. The $\mathrm{\alpha }+\mathrm{r}$ is equal to …… 

${\mathrm{C}}_1:{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-2\mathrm{y}-\mathrm{\alpha }+5=0$
$C(2, 1) r=\sqrt{4+1+\alpha -5}=\sqrt{\alpha }$
${\mathrm{C}}_2:5{\mathrm{x}}^2+5{\mathrm{y}}^2-10\mathrm{fx}-10\mathrm{gy}+36=0.$
$C_2:x^2+y^2-2fx-2gy+\frac{36}{5}=0$
$C=(f,g,)r=\sqrt{f^2+g^2-\frac{36}{5}}$
$Centre of{C}_2{\text{ is reflection of centre of C}}_1\text{ in }2x-y+1=0$

$$\begin{gathered} \begin{array}{l}\frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-1}{-1}=\frac{-2(4+1-1)}{5}=\frac{-8}{5}\\ \mathrm{x}=-\frac{6}{5};\mathrm{y}=\frac{13}{5} \\ f=-\frac{6}{5}; g=\frac{13}{5}\\ {\mathrm{r}}^2=\frac{36}{25}+\frac{169}{25}-\frac{36}{5}=1\\ \mathrm{\alpha }=1\end{array} \end{gathered}$$