(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 \(E(-1, 4)\).

(a) To find the equation of the tangent, first calculate the gradient of line AB:

\(m_{AB} = \frac{4 - 2}{-1 - 3} = -\frac{1}{2}\).

The gradient of the tangent is the negative reciprocal: \(2\).

Using point B (3, 2), the equation of the tangent is:

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

(b) The radius of the circle with centre B is the distance AB:

\(AB^2 = (4 - 2)^2 + (-1 - 3)^2 = 20\).

The equation of the circle is:

\((x - 3)^2 + (y - 2)^2 = 20\).

(c) Substitute the equation of the tangent into the circle's equation:

\((x - 3)^2 + (2x - 6)^2 = 20\).

Expanding and simplifying:

\(5x^2 - 30x + 25 = 20\).

\(5(x^2 - 6x + 1) = 0\).

\((x - 5)(x - 1) = 0\).

Thus, \(x = 5\) or \(x = 1\).

(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:

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

(a) Rewrite the circle equation in standard form: \((x-4)^2 + (y+2)^2 = 16 + 4 + 5\). The centre is \((4, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) The gradient from P (1, 2) to C (4, -2) is \(-\frac{4}{3}\). The tangent at P has a gradient of \(\frac{3}{4}\). The equation is \(y - 2 = \frac{3}{4}(x - 1)\) or \(4y = 3x + 5\).

(c) Q has the same y-coordinate as P, so \(y = 2\). Q is as far to the right of C as P is to the left, so \(x = 7\). Thus, Q is (7, 2).

(d) The gradient of the tangent at Q is \(-\frac{3}{4}\). The equation of the tangent at Q is \(y - 2 = -\frac{3}{4}(x - 7)\) or \(4y + 3x = 29\). Solving \(4y = 3x + 5\) and \(4y + 3x = 29\) gives \(T\) as \(\left(4, \frac{17}{4}\right)\).

(a) The midpoint of AB is the center of the circle C. Calculate the midpoint: \(\left(\frac{-1+7}{2}, \frac{-2+4}{2}\right) = (3, 1)\).

The radius is half the distance of AB. Calculate the distance: \(\sqrt{(7 - (-1))^2 + (4 - (-2))^2} = \sqrt{64 + 36} = 10\). Thus, the radius is 5.

The equation of the circle is \((x-3)^2 + (y-1)^2 = 5^2 = 25\).

(b) The gradient of AB is \(\frac{4 - (-2)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\). The gradient of the tangent T is the negative reciprocal: \(-\frac{4}{3}\).

The equation of the tangent at B is \(y - 4 = -\frac{4}{3}(x - 7)\). Simplifying gives \(3y + 4x = 40\).

(c) The reflection of the circle C in the line T has its center at \((11, 7)\) (since B is the midpoint of the line joining the centers).

The radius remains the same, so the new equation is \((x-11)^2 + (y-7)^2 = 25\).