JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 1)
The area of the region, inside the circle $(x-2 \sqrt{3})^2+y^2=12$ and outside the parabola $y^2=2 \sqrt{3} x$ is :
$3 \pi-8$
$6 \pi-8$
$3 \pi+8$
$6 \pi-16$
Explanation
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$\begin{aligned} & \text { Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}\right) d x \\\\ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}\right) d x\end{aligned}$
$\begin{array}{r}=2\left[\frac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\frac{12}{2} \sin ^{-1}\left(\frac{x-2 \sqrt{3}}{2 \sqrt{3}}\right)\right. \left.-\frac{\sqrt{2 \sqrt{3}} x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^{2 \sqrt{3}}\end{array}$
$\begin{aligned} & =2\{3 \pi-8\} \\\\ & =6 \pi-16 \text { sq. units. }\end{aligned}$
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