Given a circle equation, find its center and related line equations, and prove a relationship between coordinates.

Step 1: Show that M is the point $(3; -1)$ (for 10.1)
The equation of the circle is given as $x^2 + y^2 - 6x + 2y - 8 = 0$.
To find the center M, we complete the square:

$$(x^2 - 6x) + (y^2 + 2y) = 8$$ $$(x^2 - 6x + 9) + (y^2 + 2y + 1) = 8 + 9 + 1$$ $$(x - 3)^2 + (y + 1)^2 = 18$$

From this standard form $(x-a)^2 + (y-b)^2 = r^2$, the center M is $(a; b) = (3; -1)$.
Thus, M is the point $\boxed{(3; -1)}$.

Step 2: Determine the equation of MR (for 10.2)
M is $(3; -1)$ and R is $(0; -4)$.
First, calculate the gradient of MR:

$$m_{MR} = \frac{y_R - y_M}{x_R - x_M} = \frac{-4 - (-1)}{0 - 3} = \frac{-3}{-3} = 1$$

Now, use the point-slope form $y - y_1 = m(x - x_1)$ with R$(0; -4)$ and $m=1$:

$$y - (-4) = 1(x - 0)$$ $$y + 4 = x$$

The equation of MR is $\boxed{y = x - 4}$.

Step 3: Show that $q = 2 - p$ (for 10.3)
We are given that $RMN = 90^\circ$. This means the line segment MR is perpendicular to the line segment MN.
From Step 2, the gradient of MR is $m_{MR} = 1$.
Since $MR \perp MN$, the gradient of MN, $m_{MN}$, must be the negative reciprocal of $m_{MR}$:

$$m_{MN} = -\frac{1}{m_{MR}} = -\frac{1}{1} = -1$$

Now, calculate $m_{MN}$ using M$(3; -1)$ and N$(p; q)$:

$$m_{MN} = \frac{y_N - y_M}{x_N - x_M} = \frac{q - (-1)}{p - 3} = \frac{q + 1}{p - 3}$$

Set this equal to $-1$:

$$\frac{q + 1}{p - 3} = -1$$ $$q + 1 = -(p - 3)$$ $$q + 1 = -p + 3$$ $$q = -p + 3 - 1$$

Thus, $\boxed{q = 2 - p}$.

Step 4: Determine the values of $p$ and $q$ (for 10.4)
N$(p; q)$ lies on the circle $(x - 3)^2 + (y + 1)^2 = 18$. Substitute $(p; q)$ into the equation:

$$(p - 3)^2 + (q + 1)^2 = 18 \quad (*)$$

From Step 3, we have $q = 2 - p$. Substitute this into equation $(*)$:

$$(p - 3)^2 + ((2 - p) + 1)^2 = 18$$ $$(p - 3)^2 + (3 - p)^2 = 18$$

Since $(3 - p)^2 = (p - 3)^2$:

$$2(p - 3)^2 = 18$$ $$(p - 3)^2 = 9$$

Take the square root of both sides:

$$p - 3 = \pm\sqrt{9}$$ $$p - 3 = \pm 3$$

Case 1: $p - 3 = 3 \implies p = 6$.
Substitute $p=6$ into $q = 2 - p$: $q = 2 - 6 = -4$. So N is $(6; -4)$.
Case 2: $p - 3 = -3 \implies p = 0$.
Substitute $p=0$ into $q = 2 - p$: $q = 2 - 0 = 2$. So N is $(0; 2)$.
From the diagram, N is in the fourth quadrant (positive $x$, negative $y$). Therefore, the correct coordinates for N are $(6; -4)$.
The values are $\boxed{p = 6 \text{ and } q = -4}$.

Step 5: Determine the equation of the circle having centre O and passing through N (for 10.5)
The center O is $(0; 0)$.
The circle passes through N$(6; -4)$ (from Step 4).
The radius squared, $r_O^2$, is the distance from O to N squared:

$$r_O^2 = (x_N - x_O)^2 + (y_N - y_O)^2$$ $$r_O^2 = (6 - 0)^2 + (-4 - 0)^2$$ $$r_O^2 = 6^2 + (-4)^2$$ $$r_O^2 = 36 + 16 = 52$$

The equation of the circle with center O$(0; 0)$ is $x^2 + y^2 = r_O^2$:

$$\boxed{x^2 + y^2 = 52}$$

Step 6: Calculate the area of the circle centred at M (for 10.6)
From Step 1, the equation of the circle centred at M is $(x - 3)^2 + (y + 1)^2 = 18$.
The radius squared for this circle is $r^2 = 18$.
The area of a circle is given by the formula $A = \pi r^2$.

$$A = \pi (18)$$

The area of the circle centred at M is $\boxed{18\pi \text{ units}^2}$.

Step 7: Calculate the ratio $\frac{NP}{MP}$ in its simplest form (for 10.7)
M is the center $(3; -1)$. R is $(0; -4)$. N is $(6; -4)$.
The tangents drawn to the circle at R and N meet at P.
We know $MR$ and $MN$ are radii, so $MR = MN = \sqrt{18}$ (from Step 1, $r^2=18$).
Also, $RMN = 90^\circ$ (given).
A tangent to a circle is perpendicular to the radius at the point of tangency. So, $MR \perp RP$ and $MN \perp NP$. This means $\angle MRP = 90^\circ$ and $\angle MNP = 90^\circ$.
Since all four angles of quadrilateral MRNP are $90^\circ$ ($\angle RMN=90^\circ$, $\angle MRP=90^\circ$, $\angle MNP=90^\circ$, and thus $\angle RPN=90^\circ$), MRNP is a rectangle.
Because $MR = MN = \sqrt{18}$, it is a square.
In a square, all sides are equal, so $NP = MR = \sqrt{18}$.
The diagonals of a square are equal. So $MP = RN$.
Calculate the length of RN:

$$$$