The area of the region (in sq.units) bounded by the curves $x^2+y^2=16$ and $\mathrm{y}^2=6 \mathrm{x}$ is

• A

  $4 \pi+4 \sqrt{3}$
• B

  $\frac{2}{3}(4 \pi+\sqrt{3})$
• Correct answer
  C

  $\frac{4}{3}(4 \pi+\sqrt{3})$
• D

  $\frac{4 \pi+\sqrt{3}}{3}$

Hence, the correct answer is option: 'C'.