A circle with centre C has equation $x^2 + y^2 + 8x - 12y = 12$ 11 (a) Find the coordinates of C and the radius of the circle - AQA - A-Level Maths Pure - Question 11 - 2017 - Paper 1

Given that the origin (0, 0) is the midpoint of the chord PQ, we can denote the coordinates of points P and Q as (x1, y1) and (x2, y2) respectively. Since the coordinates of the midpoint are:

$\left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) = (0, 0)$

1. This implies:

   • $x1 + x2 = 0$
   • $y1 + y2 = 0$ Therefore, $x2 = -x1$ and $y2 = -y1$.

2. Now, since P and Q lie on the circle, we substitute (x1, y1) and (-x1, -y1) into the circle's equation: $(x + 4)^2 + (y - 6)^2 = 64$ for point P:

   • Substituting gives: $(-x1 + 4)^2 + (y1 - 6)^2 = 64$ which simplifies to: $x1^2 + 8x1 + 16 + y1^2 - 12y1 + 36 = 64$ leading to: $x1^2 + y1^2 + 8x1 - 12y1 + 52 = 0$.

3. Using the first equation: $OP^2 = x1^2 + y1^2 = r^2 - C^2$ and substituting in the relationship derived gives: $PQ = 2\sqrt{r^2 - \left(C\right)^2}$ leading to the final relation: $PQ = 2 \sqrt{12 - 12} = \sqrt{3},$ which shows that n = 1.