(a) The center of circle \(C_1\) is the midpoint of the diameter with end-points \((-3, -5)\) and \((7, 3)\). The midpoint is \(\left( \frac{-3+7}{2}, \frac{-5+3}{2} \right) = (2, -1)\).

The radius \(r\) is half the distance between the end-points: \(r = \frac{1}{2} \sqrt{(-3-7)^2 + (-5-3)^2} = \frac{1}{2} \sqrt{100 + 64} = \frac{1}{2} \times 12 = 6\).

The equation of circle \(C_1\) is \((x-2)^2 + (y+1)^2 = 41\).

(b) Circle \(C_2\) is obtained by translating \(C_1\) by \(\begin{pmatrix} 8 \\ 4 \end{pmatrix}\). The center of \(C_2\) is \((2+8, -1+4) = (10, 3)\).

The equation of circle \(C_2\) is \((x-10)^2 + (y-3)^2 = 41\).

(c) The gradient \(m\) of the line joining the centers \((2, -1)\) and \((10, 3)\) is \(m = \frac{3 - (-1)}{10 - 2} = \frac{4}{8} = \frac{1}{2}\).

The line \(RS\) is perpendicular to this line, so its gradient is \(-2\). Using the midpoint \((6, 1)\), the equation of \(RS\) is \(y - 1 = -2(x - 6)\), which simplifies to \(y = -2x + 13\).

(d) Substitute \(y = -2x + 13\) into the equation of \(C_1\): \((x-2)^2 + (-2x+13+1)^2 = 41\).

Expanding and simplifying gives \(x^2 - 4x + 4 + 4x^2 - 40x + 100 = 41\).

This simplifies to \(5x^2 - 60x + 159 = 0\).

(a) The gradient of \(AB\) is \(-1\) because the tangents have gradient 1 and are perpendicular to the radius. The center of the circle is \((4, -1)\). Using the point-slope form, the equation of \(AB\) is \(y + 1 = -1(x - 4)\), which simplifies to \(y = -x + 3\).

(b) Substitute \(y = -x + 3\) into the circle equation \((x-4)^2 + (y+1)^2 = 40\):

\((x-4)^2 + (-x+3+1)^2 = 40\)

\(2(x-4)^2 = 40\) or \((x-4)^2 = 20\)

\(x = 4 \pm \sqrt{20}\)

Coordinates of \(A\) are \((4 - \sqrt{20}, -1 + \sqrt{20})\).

(c) The equation of the tangent at \(A\) is found using the point \((4 - \sqrt{20}, -1 + \sqrt{20})\) and gradient 1:

\(y - (-1 + \sqrt{20}) = 1(x - (4 - \sqrt{20}))\)

\(y = x - 5 + 4\sqrt{5}\)

(a) Substitute the line equation \(y = x - 9\) into the circle equation:

\((x - 1)^2 + (x - 9 + 4)^2 = 40\)

Simplify to:

\(x^2 - 6x - 7 = 0\)

Factorize to find \(x\):

\((x + 1)(x - 7) = 0\)

Thus, \(x = -1\) or \(x = 7\).

For \(x = -1\), \(y = -1 - 9 = -10\).

For \(x = 7\), \(y = 7 - 9 = -2\).

Coordinates of intersection are \((-1, -10)\) and \((7, -2)\).

(b) The midpoint \(C\) of \(AB\) is:

\(C = \left(\frac{-1 + 7}{2}, \frac{-10 + (-2)}{2}\right) = (3, -6)\)

The radius is:

\(\sqrt{(7 - (-1))^2 + (-2 - (-10))^2} / 2 = \sqrt{32}\)

The equation of the circle with center \(C(3, -6)\) and radius \(\sqrt{32}\) is:

\((x - 3)^2 + (y + 6)^2 = 32\)

(a) Substitute \(y = \frac{1}{2}x + 6\) into the circle's equation:

\((x-a)^2 + \left(\frac{1}{2}x + 6 - 3\right)^2 = 20\)

\((x-a)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 + (3-2a)x + a^2 - 11 = 0\)

Using the quadratic formula, solve for \(a\):

\((3-2a)^2 - 4\cdot\frac{5}{4}(a^2-1) = 0\)

\(a = 4\) or \(a = -16\)

(b) For \(a = 4\), the center of the circle is \((4, 3)\) and \(P = (2, 7)\). The gradient of the normal is \(-2\).

The equation of the normal is:

\(y - 3 = -2(x - 4)\) or \(y - 7 = -2(x - 2)\)

(c) The tangents parallel to the normal have the same gradient \(-2\). Using the circle's equation:

\((x-4)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 - 10x = 0\)

\(x = 0\) or \(x = 8\)

Coordinates are \((0, 1)\) and \((8, 5)\).

Equations of tangents:

\(x + y = 1\) and \(2x + y = 21\)

(a) The equation of a circle with center (h, k) and radius r is \((x-h)^2 + (y-k)^2 = r^2\). For circle P, the center is (0, 2) and the radius is 10, so the equation is \(x^2 + (y-2)^2 = 100\).

(b) The gradient of the radius at point A is \(\frac{10-2}{6-0} = \frac{4}{3}\). The gradient of the tangent is the negative reciprocal, \(-\frac{3}{4}\). Using point-slope form, the equation of the tangent at (6, 10) is \(y - 10 = -\frac{3}{4}(x - 6)\), which simplifies to \(y = \frac{3}{4}x + \frac{29}{2}\).

(c) The center of circle Q is where the tangent meets the y-axis, which is \(\left(0, \frac{29}{2}\right)\). The radius is \(\frac{5}{2} \sqrt{5}\), so the equation is \(x^2 + \left(y - \frac{29}{2}\right)^2 = \left(\frac{5}{2} \sqrt{5}\right)^2 = 31.25\). To verify the y-coordinates of intersection, solve \(x^2 + (11-2)^2 = 100\) and \(x^2 + (11-11)^2 = 31.25\), confirming y-coordinates are 11.

(d) Substitute the tangent equation into circle Q's equation: \(x^2 + \left(\frac{3}{4}x + \frac{29}{2} - \frac{29}{2}\right)^2 = 31.25\). Solving gives \(x = \pm \sqrt{20}\). For each x, find y: \(y = \frac{3}{4}x + \frac{29}{2}\), resulting in \(y = \frac{58 \pm 3\sqrt{20}}{4}\).