Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

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

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

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then OC $⊥$ AB
Let OA = a and OC = b.

Since OC $⊥$ AB, OC bisects AB
[ $∵$ perpendicular from the centre to a chord bisects the chord].

In right $Δ$ ACO, we have

$O A^{2} = O C^{2} + A C^{2}$ [by Pythagoras' theorem]

$\Rightarrow A C = \sqrt{O A^{2} - O C^{2}} = \sqrt{a^{2} - b^{2}}$
$∴ A B = 2 A C = 2 \sqrt{a^{2} - b^{2}}$ [ $∵$ C is the midpoint of AB]

i.e., Length of the chord $A B = 2 \sqrt{a^{2} - b^{2}}$

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