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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:

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

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

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

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

answer is C.

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Detailed Solution

Given 2 concentric circles has radii as a and b.
Question Image

Here,
Tangent to smaller circle = larger circle's chord.
So,
OC bisects chord AB and is perpendicular to it.
In right tringle ACO by using Pythagoras theorem, we get

{OA}^{2} = {OC}^{2} + {CA}^{2}

a^{2} = b^{2} + {CA}^{2}

\sqrt{(a^{2} - b^{2})} = CA

AB = 2 CA

[perpendicular which is drawn at the center bisect chord]

AB = 2 \sqrt{(} a^{2} - b^{2})

Hence, the correct option is 3.

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