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

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 $$r = \sqrt {{2 \over 5}} $$

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