The area of a circle is halved when its radius is decreased by $n$ . Find its radius?

\frac{\sqrt{2} n}{\sqrt{2} - 1}

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\frac{\sqrt{3} n}{\sqrt{4} - 1}

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\frac{\sqrt{4} n}{\sqrt{2} - 1}

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\frac{\sqrt{2} n}{\sqrt{4} - 1}

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Solution

The correct option is A $\frac{\sqrt{2} n}{\sqrt{2} - 1}$
Area of circle with radius $a, A = π r^{2}$
$A \propto r^{2}$
$= \frac{A_{2}}{A_{1}} = \frac{r_{2}^{2}}{r_{1}^{2}}$
$= \frac{A_{1 / 2}}{A_{1}} = \frac{(r_{1} - n)^{2}}{r_{1}^{2}}$
$= \frac{1}{2} = {(1 - \frac{n}{r_{1}})}^{2}$
$= \frac{1}{\sqrt{2}} = 1 - \frac{n}{r_{1}} \Rightarrow \frac{n}{r_{1}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$
$= r_{1} = \frac{\sqrt{2} n}{\sqrt{2} - 1}$ = option $(A)$

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