First, calculate the radius of the circle using point A and the center (-2, 5):

\(r^2 = (7 + 2)^2 + (12 - 5)^2 = 9^2 + 7^2 = 81 + 49 = 130\)

The equation of the circle is:

\((x + 2)^2 + (y - 5)^2 = 130\)

Substitute the line equation \(y = -2x + 26\) into the circle equation:

\((x + 2)^2 + (-2x + 26 - 5)^2 = 130\)

\((x + 2)^2 + (-2x + 21)^2 = 130\)

Expand and simplify:

\((x^2 + 4x + 4) + (4x^2 - 84x + 441) = 130\)

\(5x^2 - 80x + 445 = 130\)

\(5x^2 - 80x + 315 = 0\)

Factorize:

\(5(x - 9)(x - 7) = 0\)

Thus, \(x = 9\) or \(x = 7\).

For \(x = 9\), substitute back into the line equation:

\(y = -2(9) + 26 = -18 + 26 = 8\)

Therefore, the coordinates of B are \((9, 8)\).

(a) To find the value of p such that AC and BC are equal in length, we use the distance formula:

\((5 - 2p)^2 + (p + 2)^2 = (10 - 2p)^2 + (3 - p)^2\)

Expanding both sides:

\(25 - 20p + 4p^2 + p^2 + 4p + 4 = 100 - 40p + 4p^2 + 9 - 6p + p^2\)

Simplifying gives:

\(30p = 80\)

Thus, \(p = \frac{8}{3}\).

(b)(i) Given that AC is perpendicular to BC, the slopes \(m_{AC}\) and \(m_{BC}\) must satisfy \(m_{AC} \times m_{BC} = -1\).

\(m_{AC} = \frac{p + 2}{2p - 5}\) and \(m_{BC} = \frac{p - 3}{2p - 10}\)

Setting \(\frac{p + 2}{2p - 5} \times \frac{p - 3}{2p - 10} = -1\) and solving gives:

\(p^2 - p - 6 = 0\)

Factoring gives \((p - 4)(p - 11) = 0\), so \(p = 4\) (ignoring \(p = \frac{11}{5}\) as p is an integer).

(b)(ii) The circle passes through A, B, and C. The midpoint of AB is \((7.5, 0.5)\).

The radius squared \(r^2 = \frac{50}{4}\) or \(r = \frac{5\sqrt{2}}{2}\).

The equation of the circle is \((x - 7.5)^2 + (y - 0.5)^2 = \frac{50}{4}\).

Expanding and simplifying gives:

\(x^2 + y^2 - 15x - y + 44 = 0\).

(a) To find the perpendicular bisector of AB, first find the midpoint of AB:

Midpoint = \(\left( \frac{10+5}{2}, \frac{2+(-1)}{2} \right) = \left( \frac{15}{2}, \frac{1}{2} \right)\).

Next, find the gradient of AB:

Gradient of AB = \(\frac{-1 - 2}{10 - 5} = -\frac{3}{5}\).

The gradient of the perpendicular bisector is the negative reciprocal:

Gradient = \(\frac{5}{3}\).

Using the point-slope form, the equation of the perpendicular bisector is:

\(y - \frac{1}{2} = \frac{5}{3} \left( x - \frac{15}{2} \right)\).

Simplifying gives:

\(y = \frac{5}{3}x - 12\).

(b) The equation of a circle with center (h, k) and radius r is:

\((x-h)^2 + (y-k)^2 = r^2\).

Here, the center is A (5, 2) and it passes through B (10, -1), so:

Radius = \(\sqrt{(10-5)^2 + (-1-2)^2} = \sqrt{25 + 9} = \sqrt{34}\).

The equation of the circle is:

\((x-5)^2 + (y-2)^2 = 34\).