The tangent to the circle $$C_1: x^2 + y^2 - 2x - 1 = 0$$ at the point (2, 1) cuts off a chord of length 4 from a circle $$C_2$$ whose centre is (3, -2). The radius of $$C_2$$ is:

We begin with the first circle $$C_1: x^2 + y^2 - 2x - 1 = 0$$ and the point $$P(2,\,1)$$ lying on it. In the general form $$x^2 + y^2 + 2gx + 2fy + c = 0,$$ we have $$2g = -2 \implies g = -1,$$ $$2f = 0 \implies f = 0,$$ and $$c = -1.$$

The standard tangent formula for a circle is stated as follows: for the point $$(x_1,\,y_1)$$ on the circle, the tangent is given by

$$x x_1 + y y_1 + g\,(x + x_1) + f\,(y + y_1) + c = 0.$$

Substituting $$g=-1,\;f=0,\;c=-1,\;x_1 = 2,\;y_1 = 1,$$ we obtain

$$x\cdot 2 + y\cdot 1 + (-1)\,(x + 2) + 0\,(y + 1) - 1 = 0.$$

Simplifying step by step, we have

$$2x + y - x - 2 - 1 = 0,$$

$$x + y - 3 = 0.$$

Hence the tangent line is $$L: x + y = 3.$$

Now consider the second circle $$C_2$$ whose centre is given as $$(3,\,-2).$$ Let its radius be $$R.$$ The tangent line $$L$$ meets $$C_2$$ in a chord whose length is stated to be 4.

For a circle, if a line at a perpendicular distance $$d$$ from the centre cuts the circle, the length of the chord intercepted is given by the well-known formula

$$\text{Chord length} = 2\sqrt{R^2 - d^2}.$$

We therefore need the distance from the centre $$(3,\,-2)$$ of $$C_2$$ to the line $$x + y - 3 = 0.$$ Using the distance formula for a point $$(x_0,\,y_0)$$ to the line $$ax + by + c = 0,$$ namely

$$d = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}},$$

we substitute $$a = 1,\; b = 1,\; c = -3,\; x_0 = 3,\; y_0 = -2.$$ This yields

$$d = \dfrac{|1\cdot 3 + 1\cdot(-2) - 3|}{\sqrt{1^2 + 1^2}} = \dfrac{|3 - 2 - 3|}{\sqrt{2}} = \dfrac{|-2|}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}.$$

The chord length is given to be 4, so we set

$$4 = 2\sqrt{R^2 - d^2}.$$

Dividing by 2,

$$2 = \sqrt{R^2 - d^2}.$$

Squaring both sides gives

$$4 = R^2 - d^2.$$

Substituting $$d^2 = (\sqrt{2})^2 = 2,$$ we obtain

$$4 = R^2 - 2 \quad\Longrightarrow\quad R^2 = 6.$$

Taking the positive square root (radius is positive), we have

$$R = \sqrt{6}.$$

Hence, the correct answer is Option A.