点 $P$ 和 $Q$ 均在同一个圆上，它们的坐标分别为 $(-10, 15)$ 和 $(10, 0)$。该圆的圆心在直线 $4y = 3x + 37$ 上。 点 $R$ 使得 $PR$ 是该圆的一条直径。

1. Find the coordinates of the midpoint of $PQ$.

Midpoint of $PQ = \left( \frac{-10 + 10}{2}, \frac{15 + 0}{2} \right) = \left( 0, \frac{15}{2} \right)$

2. Find the gradient of the line segment $PQ$.

Gradient of $PQ = \frac{0 - 15}{10 - (-10)} = \frac{-15}{20} = -\frac{3}{4}$

3. Determine the gradient of the perpendicular bisector.

Gradient of perpendicular bisector $= \frac{-1}{-\frac{3}{4}} = \frac{4}{3}$

4. Form the equation of the perpendicular bisector.

Equation of perpendicular bisector:
$y - \frac{15}{2} = \frac{4}{3}(x - 0)$
$y = \frac{4}{3}x + \frac{15}{2}$

1. Find the equation of the perpendicular bisector of $PQ$ (from part i)

Midpoint of $PQ = \left(\frac{-10+10}{2}, \frac{15+0}{2}\right) = (0, 7.5)$
Gradient of $PQ = \frac{0-15}{10-(-10)} = -\frac{3}{4}$
Gradient of perpendicular bisector = $\frac{4}{3}$
Equation of perpendicular bisector: $y - 7.5 = \frac{4}{3}(x - 0) \implies y = \frac{4}{3}x + 7.5$

2. Find the coordinates of the centre of the circle

The centre lies on the perpendicular bisector and the line $4y = 3x + 37$.
Substitute $y = \frac{4}{3}x + 7.5$ into $4y = 3x + 37$:
$4\left(\frac{4}{3}x + 7.5\right) = 3x + 37$
$\frac{16}{3}x + 30 = 3x + 37$
$\frac{7}{3}x = 7 \implies x = 3$
$y = \frac{4}{3}(3) + 7.5 = 11.5$
Centre of circle is $(3, 11.5)$.

3. Calculate the square of the radius

Using point $Q(10, 0)$ and centre $(3, 11.5)$:
$r^2 = (10 - 3)^2 + (0 - 11.5)^2$
$r^2 = 7^2 + (-11.5)^2 = 49 + 132.25 = 181.25$

4. Form the equation of the circle and expand to general form

Equation of circle: $(x - 3)^2 + (y - 11.5)^2 = 181.25$
$x^2 - 6x + 9 + y^2 - 23y + 132.25 = 181.25$
$x^2 + y^2 - 6x - 23y + 141.25 - 181.25 = 0$
$x^2 + y^2 - 6x - 23y - 40 = 0$

1. Find the coordinates of the centre of the circle from earlier parts

The centre $C$ is the intersection of the perpendicular bisector of $PQ$ and the line $4y = 3x + 37$.
Midpoint of $PQ = \left(\frac{-10+10}{2}, \frac{15+0}{2}\right) = (0, 7.5)$.
Gradient of $PQ = \frac{0 - 15}{10 - (-10)} = -\frac{3}{4}$.
Gradient of perpendicular bisector = $\frac{4}{3}$.
Equation of perpendicular bisector: $y - 7.5 = \frac{4}{3}(x - 0) \implies 4y = \frac{16}{3}x + 30$.
Solving simultaneously with $4y = 3x + 37$:
$\frac{16}{3}x + 30 = 3x + 37 \implies \frac{7}{3}x = 7 \implies x = 3$.
$4y = 3(3) + 37 = 46 \implies y = 11.5$.
Centre $C$ is $(3, 11.5)$.

2. Use the midpoint formula to find the coordinates of $R$

Since $PR$ is a diameter, the centre $C(3, 11.5)$ is the midpoint of $PR$.
Let the coordinates of $R$ be $(x, y)$.
$\frac{x - 10}{2} = 3 \implies x - 10 = 6 \implies x = 16$
$\frac{y + 15}{2} = 11.5 \implies y + 15 = 23 \implies y = 8$
The coordinates of $R$ are $(16, 8)$.