The image of the point (3, 5) in the line $$x - y + 1 = 0$$, lies on:

To find the image of the point $$(3, 5)$$ in the line $$x - y + 1 = 0$$, we first find the foot of the perpendicular from $$(3, 5)$$ to this line.

The line $$x - y + 1 = 0$$ has normal direction $$(1, -1)$$. The perpendicular from $$(3, 5)$$ can be parametrized as $$(3 + t, 5 - t)$$. For this point to lie on the line: $$(3 + t) - (5 - t) + 1 = 0$$, which gives $$2t - 1 = 0$$, so $$t = \frac{1}{2}$$.

The foot of the perpendicular is $$\left(\frac{7}{2}, \frac{9}{2}\right)$$.

The image is obtained by reflecting through the foot, so the image point is $$(3 + 2 \times \frac{1}{2},\; 5 - 2 \times \frac{1}{2}) = (4, 4)$$.

Now we check which circle contains $$(4, 4)$$. For the circle $$(x - 2)^2 + (y - 4)^2 = 4$$: $$(4 - 2)^2 + (4 - 4)^2 = 4 + 0 = 4$$. This is satisfied.

Therefore, the image of $$(3, 5)$$ lies on $$(x - 2)^2 + (y - 4)^2 = 4$$.