Multiple choice

The chord of circle $C$, having centre at $(2,2)$ is the diameter of the circle $x^{2}+y^{2}-4x+6y-3=0$. If $r$ is radius of the circle $C$, then $r^{2}$ is equal to

  1. $41$
  2. $31$
  3. $51$
  4. $61$
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A Correct answer
Explanation

The circle x^2 + y^2 - 4x + 6y - 3 = 0 has center (2, -3) and radius sqrt(2^2 + (-3)^2 - (-3)) = sqrt(4 + 9 + 3) = 4. The chord of circle C is the diameter of this circle, so the chord length is 2 * 4 = 8. The distance from C(2, 2) to the chord (which is the line passing through (2, -3) perpendicular to the chord) is 5. Using r^2 = (chord/2)^2 + d^2, r^2 = 4^2 + 5^2 = 16 + 25 = 41.

AI explanation

The diameter of the given circle x^2 + y^2 - 4x + 6y - 3 = 0 serves as the chord of circle C. The center of this given circle is (2, -3) and its radius squared is 2^2 + (-3)^2 - (-3) = 16, meaning its endpoints (the chord endpoints for circle C) are found by moving along the line, but a simpler approach is to use the perpendicular distance from center C. The center of circle C is at (2, 2). The distance from (2, 2) to the center of the chord (2, -3) is 5. The length of the chord (the diameter of the smaller circle) is 2 * 4 = 8, so half the chord length is 4. Using the Pythagorean theorem for circle C, r^2 = 5^2 + 4^2 = 25 + 16 = 41. The result is 41.

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