Let the mirror image of a circle $$c_1: x^2 + y^2 - 2x - 6y + \alpha = 0$$ in line $$y = x + 1$$ be $$c_2: 5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$$. If r is the radius of circle $$c_2$$, then $$\alpha + 6r^2$$ is equal to ______

The circle $$c_1: x^2 + y^2 - 2x - 6y + \alpha = 0$$ has centre $$(1, 3)$$ and radius $$r_1 = \sqrt{1 + 9 - \alpha} = \sqrt{10 - \alpha}$$.

We reflect the centre $$(1, 3)$$ in the line $$y = x + 1$$, i.e., $$x - y + 1 = 0$$. The reflection of a point $$(h, k)$$ in the line $$ax + by + c = 0$$ is given by $$\left(h - \frac{2a(ah+bk+c)}{a^2+b^2},\; k - \frac{2b(ah+bk+c)}{a^2+b^2}\right)$$. With $$a = 1, b = -1, c = 1$$:

$$ah + bk + c = 1 - 3 + 1 = -1$$, and $$a^2 + b^2 = 2$$.

The reflected centre is $$\left(1 - \frac{2(1)(-1)}{2},\; 3 - \frac{2(-1)(-1)}{2}\right) = (1 + 1,\; 3 - 1) = (2, 2)$$.

The reflected circle $$c_2$$ has the same radius as $$c_1$$ and centre $$(2, 2)$$. Its equation is $$(x-2)^2 + (y-2)^2 = r_1^2 = 10 - \alpha$$, which expands to $$x^2 + y^2 - 4x - 4y + 8 - (10-\alpha) = 0$$, i.e., $$x^2 + y^2 - 4x - 4y + \alpha - 2 = 0$$.

We are told $$c_2: 5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$$, which in standard form (dividing by 5) is $$x^2 + y^2 + 2gx + 2fy + \frac{38}{5} = 0$$.

Comparing with $$x^2 + y^2 - 4x - 4y + (\alpha - 2) = 0$$: $$2g = -4 \Rightarrow g = -2$$, $$2f = -4 \Rightarrow f = -2$$, and $$\frac{38}{5} = \alpha - 2 \Rightarrow \alpha = 2 + \frac{38}{5} = \frac{48}{5}$$.

The radius of $$c_2$$ is $$r = \sqrt{g^2 + f^2 - 38/5} = \sqrt{4 + 4 - 38/5} = \sqrt{\frac{40 - 38}{5}} = \sqrt{\frac{2}{5}}$$.

So $$r^2 = \frac{2}{5}$$ and $$\alpha + 6r^2 = \frac{48}{5} + 6\cdot\frac{2}{5} = \frac{48 + 12}{5} = \frac{60}{5} = 12$$.

Hence, the correct answer is 12.