(a) To find the perpendicular bisector, first find the midpoint of \(AB\). The midpoint is \((-1, 7)\).

Next, calculate the gradient of \(AB\):

\(m = \frac{11 - 3}{5 + 7} = \frac{8}{12} = \frac{2}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal: \(-\frac{3}{2}\).

Using the point-slope form, the equation is:

\(y - 7 = -\frac{3}{2}(x + 1)\).

Simplifying gives:

\(2(y - 7) = -3(x + 1)\)

\(2y - 14 = -3x - 3\)

\(3x + 2y = 11\).

(b) The center of the circle lies on the line \(12x - 5y = 70\) and the perpendicular bisector \(3x + 2y = 11\). Solving these simultaneously:

\(12x - 5y = 70\)

\(3x + 2y = 11\)

Solving gives \(x = 5\) and \(y = -2\).

The radius is the distance from the center \((5, -2)\) to either \((-7, 3)\) or \((5, 11)\):

\(r = \sqrt{(5 - (-7))^2 + (-2 - 3)^2} = \sqrt{12^2 + 5^2} = 13\).

The equation of the circle is:

\((x - 5)^2 + (y + 2)^2 = 169\).