Multiple choice

If two chords of lengths $2a$ each, of a circle of radius $R$, intersect each other at right angles then the distance of their point of intersection from the centre of the circle is

  1. $2\sqrt{R^2-a^2}$
  2. $\sqrt{2(R^2-a^2)}$
  3. $4\sqrt{(R^2-a^2)}$
  4. $2(R^2-a^2)$
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B Correct answer
Explanation

For a chord of length 2a, the distance from the center is sqrt(R^2 - a^2). If two such chords intersect at right angles, they form a square with the center, where the distance from the center to the intersection point is the diagonal of a rectangle with sides equal to the distances of the chords from the center. Thus, distance = sqrt((sqrt(R^2-a^2))^2 + (sqrt(R^2-a^2))^2) = sqrt(2(R^2-a^2)).

AI explanation

Let the chords AB and CD intersect at right angles at point P. The perpendicular distance from the center O to chord AB is d1 = sqrt(R^2 - a^2), and the perpendicular distance to chord CD is d2 = sqrt(R^2 - a^2). Since the chords are perpendicular, the distance OP is the hypotenuse of a right triangle with legs d1 and d2. Thus, OP = sqrt(d1^2 + d2^2) = sqrt(2(R^2 - a^2)).

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