Mathematics | VITEEE 2015 | Discussion Page

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The area of the region bounded by the curves x²+y²=9 and x+y=3 is

$\frac{9\pi}{4}+\frac{1}{2}$

$\frac{9\pi}{4}-\frac{1}{2}$

$9\left[\frac{\pi}{4}-\frac{1}{2}\right]$

$9\left[\frac{\pi}{4}+\frac{1}{2}\right]$

Option C

Area of required region

= $\frac{1}{4}$ x Area of circle - Area of $\triangle$ OAB

= $\frac{1}{4}\times\pi\times(3)^{2}-\frac{1}{2}\times 3\times3$

= $9\left[\frac{\pi}{4}-\frac{1}{2}\right]$

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