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Q.
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
EAMCETEAMCET 2010
Solution:
Intersection point of two circles
$x^{2}+y^{2}=1$ ...(i)
$(x-1)^{2} +y^{2}=1$ ...(ii)
is given by
$(x-1)^{2}+\left(1-x^{2}\right)=1$
$\Rightarrow x^{2}+1-2 x-x^{2}=0$
$\Rightarrow x=\frac{1}{2}$
From Eq. (i), $\frac{1}{4}+y^{2}=1$
$y^{2}=1-\frac{1}{4}$
$\Rightarrow y=\pm \frac{\sqrt{3}}{2}$
Point $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \text { and } C\left(\frac{1}{2}, \frac{-\sqrt{3}}{2}\right)$
So, Area of region $O A B C O$
$=2 \times$ Area of region $O A B D O$ ...(i)
Area of $O A B D O$
$=$ Area of $O A D O+$ Area of $A B D A$
$=\int\limits_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x+\int\limits_{1 / 2}^{1} \sqrt{1-x^{2}} d x$
$=\left[\frac{1}{2} \cdot(x-1) \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right]_{0}^{1 / 2}$
$+\left[\frac{1}{2} x \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_{1 / 2}^{1}$
$=\left[-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \sin ^{-1}\left(\frac{-1}{2}\right)-\frac{1}{2} \cdot 0-\frac{1}{2} \sin ^{-1}(-1)\right]$
$+\left[\frac{1}{2} \cdot 0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right]$
$=\left(-\frac{\sqrt{3}}{8}\right)-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)+\frac{1}{2} \sin ^{-1}(1)$
$+\frac{1}{2} \sin ^{-1}(1)-\frac{\sqrt{3}}{8}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)$
$=\sin ^{-1}(1)-\sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{4}$
$=\frac{\pi}{2}-\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
$=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$
from Eq. (i),
Area of region $O A B C O=2 \times\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$
$=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$
