Question

A circle centered 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.

Hint:

For questions involving distances from the center, always work with \(d^2\) instead of \(d\) to avoid carrying square roots through your algebraic steps.

Answer 1

Centre of circle: C(2, 1)
The circle passes through A(5, 6) and B(–3, K).

Step 1: Find radius using point A
Radius² = CA²
\[ CA^2 = (5 - 2)^2 + (6 - 1)^2 \] \[ = 3^2 + 5^2 = 9 + 25 = 34 \]
So, \[ R^2 = 34 \]

Step 2: Apply same radius condition for point B(–3, K)
\[ CB^2 = (-3 - 2)^2 + (K - 1)^2 \] \[ = 25 + (K - 1)^2 \] Since B lies on the circle: \[ 25 + (K - 1)^2 = 34 \] \[ (K - 1)^2 = 9 \] \[ K - 1 = \pm 3 \] \[ K = 4 \quad \text{or} \quad K = -2 \]
Values of K: \[ \boxed{4,\ -2} \]

Step 3: Find the length of chord AB
Points: A(5, 6), B(–3, K)
\[ AB^2 = (5 + 3)^2 + (6 - K)^2 \] \[ = 8^2 + (6 - K)^2 \] \[ = 64 + (6 - K)^2 \]

Case 1: K = 4
\[ AB^2 = 64 + (6 - 4)^2 = 64 + 4 = 68 \] \[ AB = \sqrt{68} = 2\sqrt{17} \]
Case 2: K = –2
\[ AB^2 = 64 + (6 + 2)^2 = 64 + 64 = 128 \] \[ AB = \sqrt{128} = 8\sqrt{2} \]

Final Answers:
Values of K: \[ \boxed{K = 4,\ K = -2} \]
Corresponding chord lengths: \[ \boxed{AB = 2\sqrt{17}\ \text{or}\ 8\sqrt{2}} \]