(a) Calculate the gradients of \(AB\) and \(BC\):

Gradient of \(AB = \frac{-3}{5}\), Gradient of \(BC = \frac{5}{3}\).

The product of the gradients \(m_{AB} \times m_{BC} = -1\), confirming \(\angle ABC = 90^\circ\).

(b) The center \(D\) is the midpoint of \(AC\):

\(D = \left( \frac{-2+6}{2}, \frac{3+5}{2} \right) = (2, 4)\).

(c) Use the circle equation \((x - x_c)^2 + (y - y_c)^2 = r^2\):

\((x-2)^2 + (y-4)^2 = 17\).

(d) Find \(E\) such that \(BE\) is a diameter:

\(E = (1, 8)\).

The gradient of the tangent is the negative reciprocal of \(BE\):

Gradient of \(BE = -4\), so gradient of tangent = \(\frac{1}{4}\).

Equation of the tangent: \(y - 8 = \frac{1}{4}(x - 1)\).