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

The slope ($$m$$) of the line $$AB$$: $$m = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$$

$$y - (-11) = \frac{3}{2}(x - 4)$$

$$2y + 22 = 3x - 12 \implies 3x - 2y - 34 = 0$$

So, the equation of the line $$AB$$ is $$3x - 2y - 34 = 0$$

The line $$AB$$ is the chord of contact of point $$C$$ with respect to the original circle: $$x^2 + y^2 - 3x + 10y - 15 = 0$$

Comparing this with the general circle equation gives $$2g = -3 \implies g = -\frac{3}{2}$$ and $$2f = 10 \implies f = 5$$.

The equation of the chord of contact from $$C(x_0, y_0)$$: $$x x_0 + y y_0 - \frac{3}{2}(x + x_0) + 5(y + y_0) - 15 = 0$$

$$\left(x_0 - \frac{3}{2}\right)x + (y_0 + 5)y - \frac{3}{2}x_0 + 5y_0 - 15 = 0$$

Since this equation represents the exact same line as $$3x - 2y - 34 = 0$$, we compare their corresponding coefficients:

$$\frac{x_0 - \frac{3}{2}}{3} = \frac{y_0 + 5}{-2} = \frac{-\frac{3}{2}x_0 + 5y_0 - 15}{-34} = k$$

$$x_0 - \frac{3}{2} = 3k \implies x_0 = 3k + \frac{3}{2}$$

$$y_0 + 5 = -2k \implies y_0 = -2k - 5$$

$$\frac{-\frac{3}{2}\left(3k + \frac{3}{2}\right) + 5(-2k - 5) - 15}{-34} = k$$

$$-\frac{9}{2}k - \frac{9}{4} - 10k - 25 - 15 = -34k$$

$$k =\frac{13}{6}$$

$$x_0 = 3\left(\frac{13}{6}\right) + \frac{3}{2} = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8$$

$$y_0 = -2\left(\frac{13}{6}\right) - 5 = -\frac{13}{3} - 5 = -\frac{28}{3}$$

So, the center point is $$C\left(8, -\frac{28}{3}\right)$$

The radius is the perpendicular distance from $$C\left(8, -\frac{28}{3}\right)$$ to the line $$3x - 2y - 34 = 0$$:

$$\text{Radius } (r) = \frac{\left|3(8) - 2\left(-\frac{28}{3}\right) - 34\right|}{\sqrt{3^2 + (-2)^2}}$$

$$r = \frac{\left|24 + \frac{56}{3} - 34\right|}{\sqrt{9 + 4}} = \frac{\left|\frac{56}{3} - 10\right|}{\sqrt{13}}$$

$$r = \frac{\left|\frac{26}{3}\right|}{\sqrt{13}} = \frac{26}{3\sqrt{13}} = \frac{2\sqrt{13}}{3}$$