(a) To show that line l is tangent to the circle at A, we need to prove that the distance from the center of the circle to the line is equal to the radius of the circle.

The equation of the circle is \((x - 5)^2 + (y - 11)^2 = 52\).

The equation of the line is \(y - 5 = -\frac{2}{3}(x - 1)\), which simplifies to \(y = -\frac{2}{3}x + \frac{17}{3}\).

Substitute \((1, 5)\) into the circle equation: \((5 - 1)^2 + (11 - 5)^2 = 52\), which holds true, confirming A is on the circle.

The gradient of the radius at A is the negative reciprocal of the line's gradient, \(\frac{3}{2}\), confirming the line is tangent.

(b) For the other circle, the center must be such that the line is tangent at A. The center is \((-3, -1)\) because the line's perpendicular distance to this center is the radius \(\sqrt{52}\).

The equation of the circle is \((x + 3)^2 + (y + 1)^2 = 52\).

(a) To find p, use the condition that line AB is perpendicular to line BC. The gradient of AB is \(\frac{7 - 4}{p - 6}\) and the gradient of BC is \(\frac{18 - 7}{14 - p}\). Since they are perpendicular, their product is -1:

\(\frac{7 - 4}{p - 6} \times \frac{18 - 7}{14 - p} = -1\)

Solving this equation gives \(p^2 - 20p + 51 = 0\). Solving the quadratic equation, we find \(p = 3\) or \(p = 17\). Given \(p < 10\), the value of \(p\) is 3.

(b) The center of the circle is the midpoint of AC, which is \((10, 11)\). The radius is the distance from the center to any of the points, say A: \(\sqrt{(10 - 6)^2 + (11 - 4)^2} = \sqrt{65}\). Thus, the equation of the circle is \((x - 10)^2 + (y - 11)^2 = 65\).

(c) The gradient of the radius at C is \(\frac{11 - 18}{10 - 14} = \frac{7}{4}\). The tangent at C is perpendicular to this radius, so its gradient is \(-\frac{4}{7}\). Using point-slope form at C (14, 18), the equation is:

\(y - 18 = -\frac{4}{7}(x - 14)\)

Rearranging gives \(4x + 7y - 182 = 0\).

(a) To find the \(x\)-coordinates where the circle intersects the \(x\)-axis, set \(y = 0\) in the circle's equation:

\(x^2 - 4x - 77 = 0\).

Factor the quadratic equation:

\((x + 7)(x - 11) = 0\).

Thus, the \(x\)-coordinates are \(-7\) and \(11\).

(b) The center of the circle \(C\) is \((2, -3)\). The gradient of \(AC\) is \(\frac{1}{3}\) or the gradient of \(BC\) is \(\frac{1}{3}\).

The gradient of the tangent at \(A\) is \(3\) or the gradient of the tangent at \(B\) is \(-3\).

The equations of the tangents are \(y = 3x + 21\) and \(y = -3x + 33\).

Set the equations equal to find the intersection:

\(3x + 21 = -3x + 33\).

Solve for \(x\):

\(6x = 12\)

\(x = 2\).

Substitute \(x = 2\) into one of the tangent equations to find \(y\):

\(y = 3(2) + 21 = 27\).

Thus, the point of intersection is \((2, 27)\).

(a) To find the equation of the circle, we first determine the center. The perpendicular bisectors of the chords \(AB\) and \(BC\) intersect at the center of the circle. The midpoint of \(AB\) is \((7, 5)\) and the midpoint of \(BC\) is \((4, 9)\). The center of the circle is \((4, 5)\).

The radius \(r\) is the distance from the center \((4, 5)\) to any of the points on the circle, such as \(A(7, 1)\):

\(r^2 = (7-4)^2 + (1-5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)

Thus, the radius \(r = 5\).

The equation of the circle is \((x-4)^2 + (y-5)^2 = 25\).

(b) The gradient of the radius at \(B\) is:

\(\text{Gradient of radius} = \frac{9-5}{7-4} = \frac{4}{3}\)

The tangent at \(B\) is perpendicular to the radius, so its gradient is the negative reciprocal:

\(\text{Gradient of tangent} = -\frac{3}{4}\)

The equation of the tangent line at \(B(7, 9)\) is:

\(y - 9 = -\frac{3}{4}(x - 7)\)

Simplifying gives:

\(y = -\frac{3}{4}x + \frac{57}{4}\)

(a) Calculate the distance from \(T(-6, 6)\) to the center \(C(8, 4)\):

\((x - 8)^2 + (y - 4)^2 = (-6 - 8)^2 + (6 - 4)^2 = 200\)

\(\sqrt{200} = 10\), hence \(\sqrt{200} > 10\), so \(T\) is outside the circle.

(b) Use the formula for the angle between a tangent and a radius:

\(\text{angle} = \sin^{-1}\left(\frac{10}{10\sqrt{2}}\right) = 45^\circ\)

(c) Find the gradient \(m\) of \(CT\):

\(m = \frac{1}{7}\)

Equation of \(AB\) is \(y - 5 = 7(x - 1)\)

\(y = 7x - 2\)

(d) Substitute \(y = 7x - 2\) into the circle's equation:

\((x - 8)^2 + (7x - 2 - 4)^2 = 100\)

\(50x^2 - 100x = 0\)

\(x = 0\) and \(x = 2\)