Mathematics — CBSE Class 10 board question

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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• 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 $\sqrt{128} = 8\sqrt{2}$.