QJEEM22Ma6deb: JEE Main 2022 Mathematics - Class 11 JEEM22Ma6deb Solution: Let the mirror image of a circle in line be . If is the radius…

${c_1}:{x^2} + {y^2} - 2x - 6y + \alpha = 0$

Then centre $= (1,3)$ and radius $(r) = \sqrt {10 - \alpha }$

Image of $(1,3)$ w.r.t. line $x - y + 1 = 0$ is $(2,2)$

${c_2}:5{x^2} + 5{y^2} + 10gx + 10fy + 38 = 0$

or ${x^2} + {y^2} + 2gx + 2fy + {{38} \over 5} = 0$

Then $( - g, - f) = (2,2)$

$\therefore$ $g = f = - 2$ .......... (i)

Radius of ${c_2} = r = \sqrt {4 + 4 - {{38} \over 5}} = \sqrt {10 - \alpha }$

$\Rightarrow {2 \over 5} = 10 - \alpha$

$\therefore$ $\alpha = {{48} \over 5}$ and $r = \sqrt {{2 \over 5}}$

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