If one of the diameters of the circle $$x^2 + y^2 - 2x - 6y + 6 = 0$$ is a chord of another circle $$C$$, whose center is at (2, 1), then its radius is ______.

The first circle is $$x^2 + y^2 - 2x - 6y + 6 = 0$$. We rewrite it in standard form by completing the square: $$(x - 1)^2 + (y - 3)^2 = 1 + 9 - 6 = 4$$.

So the center of the first circle is $$C_1 = (1, 3)$$ and its radius is $$r_1 = 2$$.

A diameter of this circle is a chord that passes through $$C_1 = (1, 3)$$ and has length $$2r_1 = 4$$, so the half-chord length is $$\ell = 2$$.

This diameter is a chord of the second circle $$C$$ whose center is $$C_2 = (2, 1)$$. Let the radius of circle $$C$$ be $$R$$.

We compute the distance from the center of $$C$$ to the center of the first circle: $$d = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1 + 4} = \sqrt{5}$$.

Since the diameter of the first circle passes through $$C_1 = (1, 3)$$, the perpendicular from $$C_2$$ to this chord meets it at $$C_1$$. This is because $$C_1$$ is the midpoint of every diameter of the first circle, and the perpendicular from the center of a circle to a chord bisects the chord. Here, the perpendicular distance from $$C_2$$ to the chord equals the distance from $$C_2$$ to the midpoint $$C_1$$ only when $$C_2$$, $$C_1$$, and the foot of the perpendicular are configured appropriately.

Actually, for any chord of circle $$C$$, if the perpendicular from $$C_2$$ to the chord has length $$d_\perp$$, then $$R^2 = d_\perp^2 + \ell^2$$. The midpoint of the chord (which is a diameter of the first circle) is $$C_1 = (1, 3)$$. The perpendicular from $$C_2$$ to the chord passes through the midpoint of the chord. So the perpendicular distance equals the distance from $$C_2$$ to the midpoint $$C_1$$, which is $$d = \sqrt{5}$$.

Applying the relation $$R^2 = d^2 + \ell^2 = 5 + 4 = 9$$, we get $$R = 3$$.

Hence, the answer is $$3$$.