The area of the region enclosed between the two circles $x^2+y^2 = 1 $ and $(x-1)^2 +y^2 =1$ is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{2}\right)$ sq. units
By symmetry I = II = III = IV
finding intersection points
$x^2+y^2=1$
$(x-1)^2+y^2=1$
$⇒x^2=(x-1)^2$
so $x=\frac{1}{2}$
so area required = $4 ×area\,I$
$=4 ×\int\limits_{1/2}^1\sqrt{1-x^2}dx$
$=4\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\right]_{1/2}^1$
$=\frac{2\pi }{3} -\frac{\sqrt{3}}{2}$ sq. units