The area of the region bounded by the curves $x^{2}+y^{2}=9$ and $x+y=3$ is

• A

  $\frac{9 \pi}{4}+\frac{1}{2}$
• B

  $\frac{9 \pi}{4}-\frac{1}{2}$
• Correct answer
  C

  $9\left(\frac{\pi}{4}-\frac{1}{2}\right)$
• D

  $9\left(\frac{\pi}{4}+\frac{1}{2}\right)$

Solution

Area of required region $\begin{array}{l} =\frac{1}{4} \times \text { Area of circle }-\text { Area of } \Delta \mathrm{OAB} \\ =\frac{1}{4} \times \pi \times(3)^{2}-\frac{1}{2} \times 3 \times 3 \\ =9\left(\frac{\pi}{4}-\frac{1}{2}\right) \end{array}$

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