Let the mirror image of a circle c _ 1 : x ^ 2 + y ^ 2 - 2x - 6y +...

Let the mirror image of a circle ${c}_{1}: {x}^{2} + {y}^{2} - 2x - 6y + \alpha = 0$ in line $y = x + 1$ be ${c}_{2}: 5{x}^{2} + 5{y}^{2} + 10gx + 10fy + 38 = 0$. If $r$ is the radius of circle ${c}_{2}$, then \alpha + 6{r}^{2} is equal to ______</p><p>Now Image of centre{c}_{1} ≡ \left(1, 3\right) in x - y + 1 = 0$is given by</p><p>$\dfrac{{x}_{1} - 1}{1} = \dfrac{{y}_{1} - 3}{- 1} = \dfrac{- 2\left(1 - 3 + 1\right)}{{1}^{2} + {1}^{2}}$</p><p>$⇒ {x}_{1} = 2, {y}_{1} = 2$</p><p>So, centre of circle${c}_{2} ≡ \left(2, 2\right)$</p><p>Now radius of circle${c}_{2}$will be$\sqrt{4 + 4 - \dfrac{38}{5}} = \sqrt{\dfrac{2}{5}} = r$</p><p>Now \left(\text{radius of}{{c}_{1}}^{2}\right) ={\left(\text{radius of}{c}_{2}\right)}^{2}$ as they are image to one another,

$⇒ \sqrt{{1}^{2} + {3}^{2} - \alpha} = \sqrt{\dfrac{2}{5}}$

$⇒ 10 - \alpha = \dfrac{2}{5} ⇒ \alpha = \dfrac{48}{5}$

Now putting the value of $\alpha \text{ and } r in \alpha + 6{r}^{2}$ we get,

$\alpha + 6{r}^{2} = \dfrac{48}{5} + \dfrac{12}{5} = 12$