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
JEE · Math · previous-year question
- A.$$\frac{1}{3}(2-12 \sqrt{3}+8 \pi)$$
- B.$$\frac{1}{3}(2-12 \sqrt{3}+6 \pi)$$
- C.$$\frac{1}{3}(4-12 \sqrt{3}+8 \pi)$$
- D.$$\frac{1}{3}(4-12 \sqrt{3}+6 \pi)$$
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