Mathematics - Area Under Curves Question with Solution | TestHub

Given circles are $x^2+y^2=1\dots$ (i) and $(x-1{)}^2+y^2=1\dots$ (ii) Centre of (i) is $\mathrm{O}(0,0)$ and radius $=1$
Both these circle are symmetrical about $x$-axis solving (i) and (ii), we get, $-2x+1=0$ $\Rightarrow x=\frac{1}{2}$ $\begin{array}{l}\text{ then }{y}^2=1-{\left(\frac{1}{2}\right)}^2=34\\ \Rightarrow y=\frac{\sqrt{3}}{2}\end{array}$ $\therefore$ The points of intersection are $\mathrm{P}\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and $\mathrm{Q}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ 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 $LAPL$ ) $=2\left[{\int }_0^{\frac{1}{2}}\sqrt{1-(x-1{)}^2}dx+{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}\mathrm{dx}\right]$ $=2{\left[\frac{(x-1)\sqrt{1-(x-1{)}^2}}{2}+\frac{1}{2}{\sin }^{-1}(x-1)\right]}_0^{\frac{1}{2}}+2{\left[\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}{\sin }^{-1}x\right]}_{\frac{1}{2}}^1$ $=-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+{\sin }^{-1}\left(\frac{-1}{2}\right)-{\sin }^{-1}(-1)+0+{\sin }^{-1}(1)-\left(\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+{\sin }^{-1}\left(\frac{1}{2}\right)\right)$ $=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$ sq. units.