The coordinates of the centre of a circle are $(x - 7, 2x)$. Find the value(s) of '$x$', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Let centre = $(x-7,\ 2x)$ and the point on the circle = $(-9, 11)$.
Since the radius = distance from centre to point = $5\sqrt{2}$:
$$\sqrt{(-9-(x-7))^2+(11-2x)^2} = 5\sqrt{2}$$
$$(-9-x+7)^2+(11-2x)^2 = 50$$
$$(-2-x)^2+(11-2x)^2 = 50$$
$$(x+2)^2+(11-2x)^2 = 50$$
$$x^2+4x+4+121-44x+4x^2 = 50$$
$$5x^2-40x+125 = 50$$
$$5x^2-40x+75 = 0$$
$$x^2-8x+15 = 0$$
$$(x-3)(x-5) = 0$$
$$\boxed{x = 3 \text{ or } x = 5}$$
Source: Distance Formula, Chapter 7
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Explanation
- Set up distance = radius using the distance formula, then square both sides to remove the square root.
- Simplify carefully to get a quadratic; factorise to get two values of $x$.
- Both values are valid — always check for two solutions in such problems.
- Examiners award marks for correct formula application, correct squaring/expansion, and correct final values.