The points $P(-3, 2)$, $Q(9, 10)$ and $R(a, 4)$ lie on the circle $C_s$, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

To show that $a = 13$, we will use the fact that the line segment $PR$ is a diameter of the circle. This implies that the midpoint of $PR$ must be the center of the circle.

1. Find the coordinates of the midpoint of $PR$: The coordinates of point $P$ are $(-3, 2)$ and those of $R$ are $(a, 4)$. The midpoint $M$ of the segment $PR$ is given by:

   $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-3 + a}{2}, \frac{2 + 4}{2} \right) = \left( \frac{-3 + a}{2}, 3 \right)$

2. Determine the center of the circle $C$: The coordinates of point $Q(9, 10)$ must also lie on the circle and are used to find a relationship:

   Using the property of the midpoint, if $O$ is the center of the circle, then:

   $\frac{-3 + a}{2} = 9 \implies -3 + a = 18 \implies a = 21$ (this is incorrect; we verify this later), but let's directly find $a$ based on $PR$'s diameter.

3. Find the slope of $PQ$ and $QR$: Using the given coordinates:

   • Slope of $PQ$ (from $P$ to $Q$): $m_{PQ} = \frac{10 - 2}{9 - (-3)} = \frac{8}{12} = \frac{2}{3}$
   • Slope of $QR$ (from $Q$ to $R$): $m_{QR} = \frac{4 - 10}{a - 9} = \frac{-6}{a - 9}$

4. Since $PR$ forms a right angle with $PQ$ at point $Q$, calculate the product of slopes:

   $m_{PQ} \cdot m_{QR} = -1 \implies \frac{2}{3} \cdot \frac{-6}{a - 9} = -1 \implies \frac{-12}{3(a - 9)} = -1 \implies 12 = 3(a - 9) \implies 12 = 3a - 27 \implies 3a = 39 \implies a = 13$