Consider a circle C _ 1 : x ^ 2 + y ^ 2 – 4x – 2y = α – 5 . Let its...

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 ${C}_{2}: 5{x}^{2} + 5{y}^{2} – 10fx – 10gy + 36 = 0$

$⇒ {C}_{2}: {x}^{2} + {y}^{2} – 2fx – 2gy + \dfrac{36}{5} = 0$

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

Also given ${O}_{2}$ is reflection of ${O}_{1}$ in $2x – y + 1 = 0$, so image formula we get,

$⇒ \dfrac{f - 2}{2} = \dfrac{g - 1}{- 1} = - 2 \cdot \left(\dfrac{2 \times 2 - 1 + 1}{{2}^{2} + {1}^{2}}\right)$

$⇒ f = \dfrac{- 6}{5}$ and $g = \dfrac{13}{5}$

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

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

Hence, $r + \alpha = 2$.