The coordinates of the centre of a circle are $(2a, a - 7)$. Find the value(s) of $a$ if the circle passes through the point $(11, -9)$ and has diameter $10\sqrt{2}$ units.
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
Diameter = $10\sqrt{2}$ units, so radius $r = 5\sqrt{2}$ units.
Since the circle passes through $(11, -9)$ and has centre $(2a,\ a-7)$:
$$\text{(radius)}^2 = (11 - 2a)^2 + (-9 - (a-7))^2 = (5\sqrt{2})^2 = 50$$
$$(11 - 2a)^2 + (-2 - a)^2 = 50$$
$$121 - 44a + 4a^2 + 4 + 4a + a^2 = 50$$
$$5a^2 - 40a + 75 = 0$$
$$a^2 - 8a + 15 = 0$$
$$(a-3)(a-5) = 0$$
$$\boxed{a = 3 \quad \text{or} \quad a = 5}$$
Source: Distance Formula, Chapter 7, Section 7.2
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Explanation
- The key idea: distance from centre to any point on the circle = radius. Square both sides to avoid surds.
- Radius = diameter ÷ 2 = $5\sqrt{2}$, so radius² = 50.
- Substitute carefully: $-9 - (a-7) = -9 - a + 7 = -(a+2)$, so the second bracket is $-(a+2)$, giving $(a+2)^2$.
- Examiners award 1 mark for correct distance equation and 1 mark for both correct values of $a$.