The area of the smaller region enclosed by the curves $y^{2}=8 x+4$ and $x^{2}+y^{2}+4 \sqrt{3} x-4=0$ is equal to

Question

$\frac{1}{3}(2-12 \sqrt{3}+8 \pi)$

$\frac{1}{3}(2-12 \sqrt{3}+6 \pi)$

$\frac{1}{3}(4-12 \sqrt{3}+8 \pi)$

$\frac{1}{3}(4-12 \sqrt{3}+6 \pi)$

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