The area of the region enclosed between two circles $x^{2} + y^{2} = 1$ and ${(x - 1)}^{2} + y^{2} = 1$ is

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$\frac{π}{3} + \frac{\sqrt{3}}{2}$

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

$\frac{π}{2} + \frac{\sqrt{3}}{2}$

$\frac{π}{2} - \frac{\sqrt{3}}{2}$

answer is C.

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Detailed Solution

$(x - 1)^{2} + y^{2} = 1, x^{2} + y^{2} = 1$

$solving (1) \& (2) - 2 x + 1 = 0; x = \frac{1}{2}$

$y^{2} = 1 - {(\frac{1}{2})}^{2} = y = \frac{\sqrt{3}}{2}$

$\begin{array}{l} 2 [\int_{0}^{1 / 2} \sqrt{1 - (x - 1)^{2}} d x + \int_{1 / 2}^{1} \sqrt{1 - x^{2}} d x] \\ 2 {[\frac{(x - 1) \sqrt{1 - (x - 1)^{2}}}{2} \frac{1}{2} \sin^{- 1} (x - 1)]}_{0}^{1 / 2} + 2 {[\frac{x \sqrt{1 - x^{2}}}{2} + \frac{1}{2} \sin^{- 1} x]}_{1 / 2}^{1} \end{array}$

$\begin{array}{l} - \frac{1}{2} - \frac{\sqrt{3}}{2} + \sin^{- 1} (- \frac{1}{2}) - \sin^{- 1} (- 1) + 0 + \sin^{- 1} (1) - (\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \sin^{- 1} \frac{1}{2}) \\ - \frac{\sqrt{3}}{4} - \frac{π}{6} - (- \frac{π}{2}) + \frac{π}{2} - \frac{\sqrt{3}}{4} - \frac{π}{6} \\ \frac{2 π}{3} - \frac{\sqrt{3}}{2} sq. unit \end{array}$