The line $$2x - y + 1 = 0$$ is a tangent to the circle at the point $$(2, 5)$$ and the centre of the circle lies on $$x - 2y = 4$$. Then, the radius of the circle is:

The tangent to the circle at point $$(2, 5)$$ is $$2x - y + 1 = 0$$. The centre of the circle lies on the line $$x - 2y = 4$$.

The radius at the point of tangency is perpendicular to the tangent line. The tangent $$2x - y + 1 = 0$$ has slope $$2$$, so the radius through $$(2, 5)$$ has slope $$-\frac{1}{2}$$.

The equation of the line through $$(2, 5)$$ with slope $$-\frac{1}{2}$$ is: $$y - 5 = -\frac{1}{2}(x - 2)$$, which simplifies to $$x + 2y = 12$$ $$-(1)$$.

The centre lies on both this line and $$x - 2y = 4$$ $$-(2)$$.

Adding $$(1)$$ and $$(2)$$: $$2x = 16$$, so $$x = 8$$. From $$(2)$$: $$8 - 2y = 4$$, so $$y = 2$$. The centre is $$(8, 2)$$.

The radius is the distance from the centre $$(8, 2)$$ to the point of tangency $$(2, 5)$$: $$r = \sqrt{(8-2)^2 + (2-5)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$$.

This matches Option A: $$3\sqrt{5}$$.