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

Step 1: Understanding the Concept:
Since the circle passes through points \(A\) and \(B\), the distance from the center \(O(2, 1)\) to point \(A\) and point \(B\) must be equal to the radius of the circle.
Thus, \(OA = OB\).
Step 2: Key Formula or Approach:
We use the distance formula between two points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\):
\[ d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \]
Step 3: Detailed Explanation:
First, calculate the radius squared (\(OA^2\)):
\[ OA^{2} = (5 - 2)^{2} + (6 - 1)^{2} \]
\[ OA^{2} = (3)^{2} + (5)^{2} = 9 + 25 = 34 \]
Now, calculate \(OB^2\) and set it equal to 34:
\[ OB^{2} = (-3 - 2)^{2} + (K - 1)^{2} = 34 \]
\[ (-5)^{2} + (K - 1)^{2} = 34 \]
\[ 25 + (K - 1)^{2} = 34 \]
\[ (K - 1)^{2} = 9 \]
Taking the square root on both sides:
\[ K - 1 = \pm 3 \]
Case 1: \(K - 1 = 3 \implies K = 4\)
Case 2: \(K - 1 = -3 \implies K = -2\)
Calculating length of chord AB:
If \(K = 4\), points are \(A(5, 6)\) and \(B(-3, 4)\):
\[ AB = \sqrt{(-3 - 5)^{2} + (4 - 6)^{2}} = \sqrt{(-8)^{2} + (-2)^{2}} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \text{ units} \]
If \(K = -2\), points are \(A(5, 6)\) and \(B(-3, -2)\):
\[ AB = \sqrt{(-3 - 5)^{2} + (-2 - 6)^{2}} = \sqrt{(-8)^{2} + (-8)^{2}} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \text{ units} \]
Step 4: Final Answer:
The values of \(K\) are 4 and -2.
The corresponding lengths of chord \(AB\) are \(2\sqrt{17}\) units and \(8\sqrt{2}\) units.