(a) To find the perpendicular bisector of AB, first find the midpoint of AB:

Midpoint = \(\left( \frac{10+5}{2}, \frac{2+(-1)}{2} \right) = \left( \frac{15}{2}, \frac{1}{2} \right)\).

Next, find the gradient of AB:

Gradient of AB = \(\frac{-1 - 2}{10 - 5} = -\frac{3}{5}\).

The gradient of the perpendicular bisector is the negative reciprocal:

Gradient = \(\frac{5}{3}\).

Using the point-slope form, the equation of the perpendicular bisector is:

\(y - \frac{1}{2} = \frac{5}{3} \left( x - \frac{15}{2} \right)\).

Simplifying gives:

\(y = \frac{5}{3}x - 12\).

(b) The equation of a circle with center (h, k) and radius r is:

\((x-h)^2 + (y-k)^2 = r^2\).

Here, the center is A (5, 2) and it passes through B (10, -1), so:

Radius = \(\sqrt{(10-5)^2 + (-1-2)^2} = \sqrt{25 + 9} = \sqrt{34}\).

The equation of the circle is:

\((x-5)^2 + (y-2)^2 = 34\).