The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

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Aicharya · Accepted Answer

Intersection point of two circles

$x^2+y^2=1$ $\ldots$ (i) $(x-1)^2+y^2=1$ $\ldots$ (ii) is given by $(x-1)^2+\left(1-x^2\right)=1$ $\Rightarrow \quad x^2+1-2 x-x^2=0$ $\Rightarrow \quad x=\frac{1}{2}$ From Eq. (i), $\frac{1}{4}+y^2=1$ $y^2=1-\frac{1}{4} \Rightarrow y= \pm \frac{\sqrt{3}}{2}$ Point $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $C\left(\frac{1}{2}, \frac{-\sqrt{3}}{2}\right)$ So, Area of region $O A B C O$ $=2 \times$ Area of region $O A B D O$ Area of $O A B D O$ $=$ Area of $O A D O+$ Area of $A B D A$ $=\int_0^{1 / 2} \sqrt{1-(x-1)^2} d x+\int_{1 / 2}^1 \sqrt{1-x^2} d x$ $=\left[\frac{1}{2} \cdot(x-1) \sqrt{1-(x-1)^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right]_0^{1 / 2}$ $+\left[\frac{1}{2} x \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_{1 / 2}^1$ $=\left[-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \sin ^{-1}\left(\frac{-1}{2}\right)-\frac{1}{2} \cdot 0-\frac{1}{2} \sin ^{-1}(-1)\right]$ $+\left[\frac{1}{2} \cdot 0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right]$ $=\left(-\frac{\sqrt{3}}{8}\right)-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)+\frac{1}{2} \sin ^{-1}(1)$ $+\frac{1}{2} \sin ^{-1}(1)-\frac{\sqrt{3}}{8}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)$ $=\sin ^{-1}(1)-\sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{4}$ $=\frac{\pi}{2}-\frac{\pi}{6}-\frac{\sqrt{3}}{4}$ $=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$ from Eq. (i), Area of region $O A B C O=2 \times\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$ $=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

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