Consider a circle $$C_1: x^2 + y^2 - 4x - 2y = \alpha - 5$$. Let its mirror image in the line $$y = 2x + 1$$ be another circle $$C_2: 5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$$. Let $$r$$ be the radius of $$C_2$$. Then $$\alpha + r$$ is equal to ______.

To solve this, we use the property that a mirror image of a circle has the same radius as the original.

Rewrite $$C_1$$: $$x^2 + y^2 - 4x - 2y + 5 - \alpha = 0$$.

The center is $$(2, 1)$$ and the radius squared ($$r^2$$) is:

$$r^2 = g^2 + f^2 - c = 2^2 + 1^2 - (5 - \alpha) = 5 - 5 + \alpha = /alpha$$

So, $$r = \sqrt{\alpha}$$.

Divide the $$C_2$$ equation by 5: $$x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$$.

The radius squared for $$C_2$$ is:

$$r^2 = f^2 + g^2 - \frac{36}{5}$$

The center of $$C_2$$ $$(f, g)$$ is the mirror image of the center of $$C_1$$ $$(2, 1)$$ in the line $$2x - y + 1 = 0$$.

Using the image formula: $$\frac{f-2}{2} = \frac{g-1}{-1} = -2\frac{2(2) - 1(1) + 1}{2^2 + (-1)^2}$$

$$\frac{f-2}{2} = \frac{g-1}{-1} = -2\frac{4}{5} = -1.6$$

• $$f = 2 + 2(-1.6) = \mathbf{-1.2}$$
• $$g = 1 - 1(-1.6) = \mathbf{2.6}$$

Since radii are equal:

$$\alpha = f^2 + g^2 - \frac{36}{5} = (-1.2)^2 + (2.6)^2 - 7.2$$

$$\alpha = 1.44 + 6.76 - 7.2 = \mathbf{1}$$

Since $$r = \sqrt{\alpha}$$, then $$r = 1$$.

$$\alpha + r = 1 + 1 = \mathbf{2}$$