Let the tangents at the points $$A(4, -11)$$ and $$B(8, -5)$$ on the circle $$x^2 + y^2 - 3x + 10y - 15 = 0$$, intersect at the point $$C$$. Then the radius of the circle, whose centre is $$C$$ and the line joining $$A$$ and $$B$$ is its tangent, is equal to

A

$$\frac{3\sqrt{3}}{4}$$

B

$$2\sqrt{13}$$

C

$$\sqrt{13}$$

D

$$\frac{2\sqrt{13}}{3}$$