Question

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

• A. $\frac{2\pi }{3}+\frac{\sqrt{3}}{2}$
• B. $\frac{\pi }{3}+\frac{\sqrt{3}}{2}$
• C. $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
• D. $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$

Answer 1

Answer: (D)

Intersection point of two circles

$x^2+y^2=1$ ...(i)
$(x-1{)}^2+y^2=1$ ...(ii)
is given by
$(x-1{)}^2+\left(1-x^2\right)=1$
$\Rightarrow x^2+1-2x-x^2=0$
$\Rightarrow x=\frac{1}{2}$
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 $OABCO$
$=2\times$ Area of region $OABDO$ ...(i)
Area of $OABDO$
$=$ Area of $OADO+$ Area of $ABDA$
$=\underset{0}{\overset{1/2}{\int }}\sqrt{1-(x-1{)}^2}dx+\underset{1/2}{\overset{1}{\int }}\sqrt{1-x^2}dx$
$={\left[\frac{1}{2}\cdot (x-1)\sqrt{1-(x-1{)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{x-1}{1}\right)\right]}_0^{1/2}$
$+{\left[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{x}{1}\right)\right]}_{1/2}^1$
$=\left[-\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}{\sin }^{-1}\left(\frac{-1}{2}\right)-\frac{1}{2}\cdot 0-\frac{1}{2}{\sin }^{-1}(-1)\right]$
$+\left[\frac{1}{2}\cdot 0+\frac{1}{2}{\sin }^{-1}(1)-\frac{1}{4}\cdot \frac{\sqrt{3}}{2}-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)\right]$
$=\left(-\frac{\sqrt{3}}{8}\right)-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)+\frac{1}{2}{\sin }^{-1}(1)$
$+\frac{1}{2}{\sin }^{-1}(1)-\frac{\sqrt{3}}{8}-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)$
$={\sin }^{-1}(1)-{\sin }^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{4}$
$=\frac{\pi }{2}-\frac{\pi }{6}-\frac{\sqrt{3}}{4}$
$=\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right)$
from Eq. (i),
Area of region $OABCO=2\times \left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right)$
$=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$