Question

The area bounded by curves $(x-1{)}^2+y^2=1$ and $x^2+y^2=1$ is

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

Answer 1

Answer: (A)

Given circles are $x^2+y^2=1\dots$ (i)
and $(x-1{)}^2+y^2=1\dots$ (ii)
Centre of (i) is $O(0,0)$ and radius $=1$

Both these circle are symmetrical about $x$-axis
solving (i) and (ii), we get,
$-2x+1=0$
$\Rightarrow x=\frac{1}{2}$
then $y^2=1-{\left(\frac{1}{2}\right)}^2=34$
$\Rightarrow y=\frac{\sqrt{3}}{2}$
$\therefore$ The points of intersection are
$P\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and $Q\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
It is clear from the figure that the shaded portion in region whose area is required.
$\therefore$ Required area = area $OQAPO$
$=2\times$ area of the region $OLAP$
$=2\times ($ area of the region $OLPO+$ area of $LAPL)$
$=2\left[\underset{0}{\overset{\frac{1}{2}}{\int }}\sqrt{1-(x-1{)}^2}dx+\underset{\frac{1}{2}}{\overset{1}{\int }}\sqrt{1-x^2}dx\right]$
$=2{\left[\frac{(x-1)\sqrt{1-(x-1{)}^2}}{2}+\frac{1}{2}{\sin }^{-1}(x-1)\right]}_0^{\frac{1}{2}}+2{\left[\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}{\sin }^{-1}x\right]}_{\frac{1}{2}}^1$
$=-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+{\sin }^{-1}\left(\frac{-1}{2}\right)-{\sin }^{-1}(-1)+0+{\sin }^{-1}(1)-\left(\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+{\sin }^{-1}\left(\frac{1}{2}\right)\right)$
$=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$ sq.units .