Question
Area enclosed by the circle $x^2+y^2=a^2$ is equal to
✓
Answer
$(b) :$ We have, $x^2+y^2=a^2$, which is a circle with centre $(0,0)$ and radius $a$.
$\therefore $ Required area $=4 \times$ Area in the first quadrant
$=4 \int_0^a \sqrt{a^2-x^2} d x$
$=4\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a$
$=4\left(\frac{a^2}{2}\right) \frac{\pi}{2}=\pi a^2 \text { sq. units }$
$\therefore $ Required area $=4 \times$ Area in the first quadrant
$=4 \int_0^a \sqrt{a^2-x^2} d x$
$=4\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a$
$=4\left(\frac{a^2}{2}\right) \frac{\pi}{2}=\pi a^2 \text { sq. units }$
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