Question

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. $\frac{9\pi }{4}-\frac{1}{2}$
• C. $9(\frac{\pi }{4}-\frac{1}{2})$
• D. $9(\frac{\pi }{4}+\frac{1}{2})$

Answer 1

The correct option is C $9(\frac{\pi }{4}-\frac{1}{2})$

Area of required region
$=\frac{1}{4}\times \text{Area of circle}-\text{Area of}△OAB$
$=\frac{1}{4}\times \pi \times (3{)}^2-\frac{1}{2}\times 3\times 3$
$=9(\frac{\pi }{4}-\frac{1}{2})$

Hence, option C is correct.