The equation of a circle, centre $C$, is $x^2 + y^2 + px + \left(\frac{p}{2} + 4\right)y + k = 0$, where $p$ and $k$ are constants. It is given that $C$ lies on the line $3x - 2y - 8 = 0$.

1. Express the coordinates of the centre $C$ in terms of $p$.

Comparing $x^2 + y^2 + px + \left(\frac{p}{2} + 4\right)y + k = 0$ with the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$:
$2g = p \implies g = \frac{p}{2}$
$2f = \frac{p}{2} + 4 \implies f = \frac{p}{4} + 2$
The centre $C$ is $(-g, -f)$, so $C = \left(-\frac{p}{2}, -\frac{p}{4} - 2\right)$.

2. Substitute the coordinates of $C$ into the equation of the line to solve for $p$.

Since $C$ lies on $3x - 2y - 8 = 0$:
$3\left(-\frac{p}{2}\right) - 2\left(-\frac{p}{4} - 2\right) - 8 = 0$
$-\frac{3p}{2} + \frac{p}{2} + 4 - 8 = 0$
$-p - 4 = 0$
$p = -4$ (shown)

1. Substitute $p = -4$ to find the coefficients of $x$ and $y$

From (i), $p = -4$.
The equation of the circle is $x^2 + y^2 - 4x + \left(\frac{-4}{2} + 4\right)y + k = 0$
$x^2 + y^2 - 4x + 2y + k = 0$

2. Find the coordinates of the centre $C$

Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, the centre is $(-g, -f)$.
$2g = -4 \implies g = -2$
$2f = 2 \implies f = 1$
Therefore, the coordinates of $C$ are $(2, -1)$.

1. Find the coordinates of the centre $C$ and the radius $r$ in terms of $k$

From (i), $p = -4$. The equation of the circle is $x^2 + y^2 - 4x + 2y + k = 0$.
The centre $C$ is $(2, -1)$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (1)^2 - k} = \sqrt{5 - k}$.

2. Relate the radius to the distance from the centre to the tangent $x = -8$

Since $x = -8$ is a vertical tangent to the circle, the distance from the centre $C(2, -1)$ to the line $x = -8$ is equal to the radius $r$.
Distance $= |2 - (-8)| = 10$.
So, $r = 10$.

3. Solve for $k$

$\sqrt{5 - k} = 10$
$5 - k = 100$
$k = -95$

1. Calculate the square of the distance from the centre $C$ to point $A$

From earlier parts, the centre of the circle is $C(2, -1)$ and the radius is $r = 10$.

$CA^2 = (14 - 2)^2 + (-8 - (-1))^2$
$CA^2 = 12^2 + (-7)^2$
$CA^2 = 144 + 49 = 193$

2. Compare the distance with the radius to show $A$ is outside the circle

The square of the radius is $r^2 = 10^2 = 100$.
Since $CA^2 = 193 > 100$, we have $CA > r$.
Therefore, point $A$ lies outside the circle.

1. State the geometrical relationship between the point outside the circle, the centre, and the furthest point on the circle

For a point $X$ on the circle to be furthest from a point $A$ outside the circle, the line segment $AX$ must pass through the centre $C$ of the circle. Therefore, the points $A$, $C$, and $X$ lie on the same straight line.