Let $$C$$ be a circle passing through the points $$A(2,-1)$$ and $$B(3,4)$$. The line segment $$AB$$ is not a diameter of $$C$$. If $$r$$ is the radius of $$C$$ and its centre lies on the circle $$(x-5)^2 + (y-1)^2 = \frac{13}{2}$$, then $$r^2$$ is equal to

We need to find $$r^2$$ where $$r$$ is the radius of circle $$C$$ passing through $$A(2, -1)$$ and $$B(3, 4)$$, with centre on the circle $$(x-5)^2 + (y-1)^2 = \frac{13}{2}$$.

Since the perpendicular bisector of $$AB$$ must pass through its midpoint, the midpoint is $$\left(\frac{2+3}{2}, \frac{-1+4}{2}\right) = \left(\frac{5}{2}, \frac{3}{2}\right)$$. The slope of $$AB$$ is $$\frac{4-(-1)}{3-2} = 5$$, so the slope of the perpendicular bisector is $$-\frac{1}{5}$$. Now the equation of this bisector is $$y - \frac{3}{2} = -\frac{1}{5}\left(x - \frac{5}{2}\right),$$ which simplifies to $$y = -\frac{x}{5} + 2.$$

Next, substituting $$y = -\frac{x}{5} + 2$$ into $$(x-5)^2 + (y-1)^2 = \frac{13}{2}$$ gives $$(x-5)^2 + \left(-\frac{x}{5} + 1\right)^2 = \frac{13}{2},$$ which simplifies to $$(x-5)^2 + \frac{(x-5)^2}{25} = \frac{13}{2},$$ so that $$\frac{26(x-5)^2}{25} = \frac{13}{2}$$ and hence $$(x-5)^2 = \frac{25}{4},$$ leading to $$x - 5 = \pm\frac{5}{2}$$ and therefore $$x = \frac{15}{2} \text{ or } x = \frac{5}{2}.$$

From these values, when $$x = \frac{15}{2}$$ one obtains $$y = -\frac{15}{10} + 2 = \frac{1}{2}$$, giving the centre $$\left(\frac{15}{2}, \frac{1}{2}\right)$$. When $$x = \frac{5}{2}$$ one finds $$y = -\frac{5}{10} + 2 = \frac{3}{2}$$, giving the centre $$\left(\frac{5}{2}, \frac{3}{2}\right)$$.

However, the point $$\left(\frac{5}{2}, \frac{3}{2}\right)$$ coincides with the midpoint of $$AB$$, implying that $$AB$$ would be a diameter, contrary to the problem statement, and is therefore rejected.

Thus the centre of the required circle is $$\left(\frac{15}{2}, \frac{1}{2}\right)$$. Substituting this and the point $$A(2, -1)$$ into the distance formula yields $$r^2 = \left(\frac{15}{2} - 2\right)^2 + \left(\frac{1}{2} + 1\right)^2 = \left(\frac{11}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}.$$

Therefore, $$r^2 = \frac{65}{2}$$.

The correct answer is Option B.