Consider a circle $C_{1}: x^{2}+y^{2}-4 x-2 y=\alpha-5$. Let its mirror image in the line $y=2 x+1$ be another circle $C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0$. Let $r$ be the radius of $C_{2}$. Then $\alpha+r$ is equal to _________.

We have, <br/><br/>$$ \begin{aligned} & C_1: x^2+y^2-4 x-2 y=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^5-5=\alpha-5 \\\\ & C_1:(x-2)^2+(y-1)^2=(\sqrt{\alpha})^2 \end{aligned} $$ <br/><br/>So, centre and radius of $C_1$ are $(2,1)$ and $\sqrt{\alpha}$ respectively <br/><br/>Now, image of $(2,1)$ along the line $y=2 x+1$ is, <br/><br/>$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{2^2+(-1)^2}$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \\\\ & \Rightarrow x=\frac{-6}{5} \text { and } y=\frac{13}{5} \end{aligned} $$ <br/><br/>Now, $\left(\frac{-6}{5}, \frac{13}{5}\right)$ will be the centre of $C_2$ <br/><br/>$\therefore f=\frac{6}{5} \text { and } g=\frac{-13}{5}$ <br/><br/>Now, radius of $\mathrm{C}_2=r=\sqrt{f^2+g^2-\frac{36}{5}}$ <br/><br/>$$ \begin{aligned} & \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \\\\ & \because r=1 \text { so, } \alpha=1 \\\\ & \therefore \alpha+r=1+1=2 \end{aligned} $$ <br/><br/><b>Concept :</b> <br/><br/>Image of a point $\left(x_1, y_1\right)$ w.r.t. $a x+b y+c=0$ is $(x, y)$, then <br/><br/>$$ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\left(a^2+b^2\right)} $$