Multiple choice

Two perpendicular chords intersect in a circle. The segments of one chord are 3 and 4; the segments of the other are 6 and 2. Then the diameter of the circle is:

  1. $\sqrt{89}$
  2. $\sqrt{56}$
  3. $\sqrt{65}$
  4. $\sqrt{75}$
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C Correct answer
Explanation

For perpendicular chords intersecting at P, if segments are a, b and c, d, the radius R is given by R^2 = (a^2+b^2+c^2+d^2)/4. R^2 = (3^2+4^2+6^2+2^2)/4 = (9+16+36+4)/4 = 65/4. Diameter = 2R = 2 * sqrt(65/4) = sqrt(65).

AI explanation

Using the property of intersecting chords in a circle, the center of the circle is found by moving from the first chord's midpoint (1/2 unit from the intersection) perpendicularly by the half-distance of the second chord (2 units), and vice versa. This places the center at coordinate (8/5, 14/5) relative to the intersection point. The squared distance from the center to any endpoint, such as the one at (3, 0), is calculated as 65/4. The diameter is twice the square root of this value, giving a final result of the square root of 65.

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