A circle with area $A_{1}$ is contained in the interior of a larger circle with area $A_{1} + A_{2}$ . If the radius of the larger circle is $3$ and $A_{1}, A_{2}, A_{1} + A_{2}$ are in A.P., what is the radius of the smaller circle?

\frac{\sqrt{3}}{2}

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\frac{2}{\sqrt{3}}

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\sqrt{3}

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\frac{3}{2}

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Solution

The correct option is B $\sqrt{3}$

Area of small circle $= A_{1}$
Area of larger circle $= A_{1} + A_{2}$
Given that
$(A_{1}, A_{2}, A_{1} + A_{2})$ form an AP
Common difference
$d = A_{2} - A_{1} = A_{2} + A_{1} - A_{2} A_{2} - A_{1} = A_{1} A_{2} = 2 A_{1} (1)$

Radius of larger circle $= 3$ units
Area of larger circle $= π (3)^{2} = 9 π$
$A_{1} + A_{2} = 9 π$
From $(1)$
$2 A_{1} + A_{1} = 9 π 3 A_{1} = 9 π A_{1} = 3 π π r^{2} = 3 π$
$r = \sqrt{3}$ units

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