The points of intersection of the line $$ax + by = 0$$, ($$a \neq b$$) and the circle $$x^2 + y^2 - 2x = 0$$ are $$A(\alpha, 0)$$ and $$B(1, \beta)$$. The image of the circle with $$AB$$ as a diameter in the line $$x + y + 2 = 0$$ is:

The circle $$x^2 + y^2 - 2x = 0$$ can be written as $$(x-1)^2 + y^2 = 1$$, with center $$(1, 0)$$ and radius $$1$$.

Point $$A(\alpha,0)$$ lies on the circle, so $$\alpha^2 - 2\alpha = 0$$ and thus $$\alpha(\alpha - 2) = 0$$ gives $$\alpha = 0$$ or $$\alpha = 2$$. Since $$A$$ also lies on $$ax + by = 0$$, we have $$a\alpha = 0$$. If $$a = 0$$ then the line is $$y = 0$$, intersecting the circle at $$(0,0)$$ and $$(2,0)$$, but then a point $$B(1,\beta)$$ with $$\beta \neq 0$$ could not lie on this line. Therefore $$a \neq 0$$, which forces $$\alpha = 0$$ and gives $$A = (0,0)$$.

Point $$B(1,\beta)$$ lies on the circle, so $$1 + \beta^2 - 2 = 0$$ and hence $$\beta^2 = 1$$ leading to $$\beta = \pm 1$$. It also lies on $$ax + by = 0$$, so $$a + b\beta = 0$$ and $$\beta = -a/b$$. If $$\beta = 1$$ then $$a = -b$$, which is allowed since $$a \neq b$$; if $$\beta = -1$$ then $$a = b$$, contradicting $$a \neq b$$. Thus $$\beta = 1$$ and $$B = (1,1)$$.

With $$A = (0,0)$$ and $$B = (1,1)$$, the midpoint of $$AB$$ is $$\left(\frac{0+1}{2}, \frac{0+1}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right)$$ and the square of the radius of the circle with $$AB$$ as diameter is $$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2}$$.

To reflect the center across the line $$x + y + 2 = 0$$, we use the formula for reflecting $$(x_0,y_0)$$ across $$ax + by + c = 0$$:
$$x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}, \quad y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}.$$
Here $$a = 1$$, $$b = 1$$, $$c = 2$$, and $$(x_0,y_0) = \left(\frac{1}{2}, \frac{1}{2}\right)$$. Then
$$ax_0 + by_0 + c = \frac{1}{2} + \frac{1}{2} + 2 = 3,$$
$$x' = \frac{1}{2} - \frac{2(1)(3)}{2} = \frac{1}{2} - 3 = -\frac{5}{2},$$
$$y' = \frac{1}{2} - \frac{2(1)(3)}{2} = \frac{1}{2} - 3 = -\frac{5}{2}.$$
Thus the reflected center is $$\left(-\frac{5}{2}, -\frac{5}{2}\right)$$.

Reflection preserves distances, so the radius squared of the image circle remains $$\frac{1}{2}$$. Its equation is
$$\left(x + \frac{5}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{1}{2},$$
which expands to
$$x^2 + 5x + \frac{25}{4} + y^2 + 5y + \frac{25}{4} = \frac{1}{2},$$
then to
$$x^2 + y^2 + 5x + 5y + \frac{50}{4} - \frac{2}{4} = 0,$$
and finally to
$$x^2 + y^2 + 5x + 5y + 12 = 0.$$
Therefore the correct answer is Option A: $$x^2 + y^2 + 5x + 5y + 12 = 0.$$