Multiple choice

Two concentric circles of radii a and b, where a > b, are given. The length of the chord of the larger circle which touches the smaller circle is :

  1. $\sqrt{a^{2}-b^{2}}$
  2. $\sqrt{a^{2}+b^{2}}$
  3. $2\sqrt{a^{2}-b^{2}}$
  4. $2\sqrt{a^{2}+b^{2}}$
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C Correct answer
Explanation

The chord of the larger circle tangent to the smaller circle forms a right-angled triangle with the radius of the smaller circle (b) and the radius of the larger circle (a). By Pythagoras theorem, half the chord length is sqrt(a^2 - b^2), so the full length is 2*sqrt(a^2 - b^2).

AI explanation

Let the chord of the larger circle touch the smaller circle at point P. The radius of the smaller circle, b, is perpendicular to the chord at point P, meaning it bisects the chord. This forms a right-angled triangle where the hypotenuse is the radius of the larger circle, a, the perpendicular is b, and the base is half the chord length. Using the Pythagoras theorem, half the chord length is the square root of a squared minus b squared. The full length of the chord is twice this value, which is 2 times the square root of a squared minus b squared.

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