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Q.
If the area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}-4 x=0$, the parabola $y^{2}=4 x$ and the $X$ - axis is $A , 6 A-9 \sqrt{3}=$
TS EAMCET 2020
Solution:
Given curve,
$x^{2}+y^{2}-4 x=0 $
$\Rightarrow (x-2)^{2}+y^{2}=4$
and $y^{2}=x$ intersecting point of circle and parabola is $(0,0)$ and $(3, \sqrt{3})$
Required area
$A=\left[\frac{2}{3}(x)^{3 / 2}\right]_{0}^{3}+\left[\frac{x-2}{2} \sqrt{4-(x-2)^{2}}+2 \sin ^{-1} \frac{x-2}{2}\right]_{3}^{4}$
$A=2 \sqrt{3}+\left[\left(0+2 \sin ^{-1}(1)\right)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)\right)\right]$
$A=2 \sqrt{3}+\pi-\frac{\sqrt{3}}{2}-\frac{\pi}{3}=\frac{3 \sqrt{3}}{2}+\frac{2 \pi}{3}$
$6 A-9 \sqrt{3}=4 \pi$
