Question

The area of the region enclosed between two circles $x^2+y^2=1$ and $(x–1{)}^2+y^2=1$ is

• A. $(\frac{\pi }{2}-\frac{\sqrt{3}}{2})$
• B. $(\frac{\pi }{3}+\frac{\sqrt{3}}{2})$
• C. $(\frac{\pi }{6}-\frac{\sqrt{3}}{2})$
  
• D. $(\frac{\pi }{2}+\frac{\sqrt{3}}{2})$

Answer 1

The correct option is B $(\frac{\pi }{3}+\frac{\sqrt{3}}{2})$



$x^2+y^2=1\to \left(1\right)(x-1{)}^2+y^2=1\to (2)$
(1) is a circle with centre at (0, 0) and radius 1.
(2) is a circle with centre at (1, 0) and radius 1 and passing through (0, 0)
By solving (1) and (2); the points of intersection are
$(\frac{1}{2},\frac{\sqrt{3}}{2}),(\frac{1}{2}-\frac{\sqrt{3}}{2})$
Required area = $4{\int }_0^{\frac{1}{2}}\sqrt{1-x^2}dx=4{[\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}si{n}^{-1}x]}_0^{\frac{1}{2}}$
$=4[\frac{\sqrt{3}}{8}+\frac{\pi }{12}]=\frac{\sqrt{3}}{2}+\frac{\pi }{3}$ sq. unit