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 [27-Jul-2022-Shift-1]

Correct Answer is : 13(4−12√3+8π)

\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)

JEE Mains 27-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 9 of 90

Marks: +1, -0

The area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+4√3x−4=0 is equal to

[27-Jul-2022-Shift-1]

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