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 :

\sqrt{a^{2} - b^{2}}

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\sqrt{a^{2} + b^{2}}

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2 \sqrt{a^{2} - b^{2}}

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2 \sqrt{a^{2} + b^{2}}

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Solution

The correct option is D $2 \sqrt{a^{2} - b^{2}}$

Chord of larger circle will be the tangent to smaller circle.

Thus, OC is perpendicular to chord AB and bisects it.

By Pythagoras theorem, in right triangle ACO,

${O A}^{2} = {O C}^{2} + {C A}^{2}$
$a^{2} = b^{2} + {C A}^{2}$
$\sqrt{(a^{2} - b^{2})} = C A$
$A B = 2 C A$ [perpendicular drawn from centre bisect the chord]
$A B = 2 \sqrt{(} a^{2} - b^{2})$
option C will be the answer.

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