A circle centred at $(2, 1)$ passes through the points $A(5, 6)$ and $B(-3, K)$. Find the value(s) of $K$. Hence find length of chord $AB$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding rag
Model Answer
Step 1: Find the radius using point A(5, 6) and centre (2, 1).
$$r = \sqrt{(5-2)^2 + (6-1)^2} = \sqrt{9 + 25} = \sqrt{34}$$
Step 2: Since B(–3, K) also lies on the circle, its distance from centre = r.
$$\sqrt{(-3-2)^2 + (K-1)^2} = \sqrt{34}$$
$$25 + (K-1)^2 = 34$$
$$(K-1)^2 = 9 \implies K-1 = \pm 3$$
$$\boxed{K = 4 \text{ or } K = -2}$$
Step 3: Find length of chord AB.
Taking K = 4, B = (–3, 4):
$$AB = \sqrt{(5-(-3))^2 + (6-4)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \text{ units}$$
Taking K = –2, B = (–3, –2):
$$AB = \sqrt{(5-(-3))^2 + (6-(-2))^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \text{ units}$$
Source: Chapter 7, Section 7.2 (Distance Formula)
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Explanation
- The key idea: every point on a circle is equidistant from the centre (equal to radius). So set up the distance equation for both A and B.
- Find radius from the given point A first, then equate the distance of B to that radius.
- Solving $(K-1)^2 = 9$ gives two values of K — don't forget the negative root.
- For chord AB, substitute each value of K and apply the distance formula. Examiners expect both cases addressed.
- Simplify surds: $\sqrt{68} = 2\sqrt{17}$ and $\sqrt{128} = 8\sqrt{2}$.