Question

The area bounded by the circle $ x^2+y^2 = 4 $ and the line $ x = y\sqrt {3} $ in the first quadrant (in sq units) is

• A. $ \pi $
• B. $ \frac{\pi}{2} $
• C. $ \frac{\pi}{3} $
• D. None of these

Answer 1

Answer: (C)

We have, circle $x^2 + y^2 = 4\,\,\,...(i)$
and line, $x = y\sqrt{3}$
$\Rightarrow y = \frac{x}{\sqrt{3}}\,\,\,...(ii)$

On putting the value of $y$ from Eq, $(ii)$ to Eq. $(i)$, we get
$x^2 + (\frac{x}{\sqrt{3}})^2 = 4$
$\Rightarrow x^2 + \frac{x^2}{3} = 4$
$\Rightarrow 3x^2 + x^2 = 12$
$\Rightarrow 4x^2 = 12$
$\Rightarrow x^2 = 3$
$\Rightarrow x = \pm \sqrt{3}$
So, coordinates of $B$ are $(\sqrt{3}, 1)$. Also, circle cut the $X$-axis at $A(2,0)$ and $Y$-axis at $(0, 2)$
$[\because 2$ is radius of circle]
Required area $=$ Area of region $ODBO$
$+$ Area of region $DABD$
$=\int\limits_{0}^{\sqrt{3}} y $ (line) $dx + \int\limits_{\sqrt{3}}^{2} y$(circle) $dx$
$= \int\limits _0^{\sqrt{3}} \frac{x}{\sqrt{3}} dx + \int\limits_{\sqrt{3}}^{2} \sqrt{4 - x^2} dx$
$ = \frac{1}{\sqrt{3}} \left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}} + \int\limits_{\sqrt{3}}^{2} \sqrt{2^{2}-x^{2}} dx $
$ = \frac{1}{\sqrt{3}}\left[\frac{\left(\sqrt{3}\right)^{2}-0}{2}\right]+\left[\frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2} $
$= \frac{1}{\sqrt{3}}\times\frac{3}{2} + [\frac{2}{2}\sqrt{4-4} + 2 \sin^{-1}\left(\frac{2}{2}\right) $
$-\frac{\sqrt{3}}{2}\sqrt{4-3} -2 \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)] $
$= \frac{\sqrt{3}}{2}+0 + 2\sin^{-1}\left(1\right) - \frac{\sqrt{3}}{2} -2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$ = 2\times\frac{\pi}{2} - 2\times\frac{\pi}{3} = \frac{\pi}{3}$