(a) The radius \(r\) of the circle can be found using the distance formula between \(B(5, 1)\) and \(A(-1, -2)\):

\(r = \sqrt{(5 - (-1))^2 + (1 - (-2))^2} = \sqrt{6^2 + 3^2} = \sqrt{45}\)

The equation of the circle is \((x - 5)^2 + (y - 1)^2 = 45\).

(b) Point \(C\) is such that \(AC\) is a diameter, so \(C\) is at \((11, 4)\). The gradient of \(CD\) is \(-2\), which is perpendicular to the radius \(BC\) with gradient \(0.5\). Therefore, \(DC\) is a tangent.

(c) The other tangent from \(D\) touches the circle at \(E\). Using the equation of the tangent \(y = 2x + 6\), we find \(E(-1, 4)\).