Question

Find the area of the region enclosed between the two circles $x^2+y^2=1$ & $(x-1{)}^2+y^2=1$

• A. $\frac{\pi }{6}-\frac{\sqrt{3}}{2}$ sq.units
• B. $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$ sq.units
• C. $\frac{\pi }{6}-\frac{\sqrt{3}}{4}$ sq.units
• D. $\frac{\pi }{3}-\frac{\sqrt{3}}{4}$ sq.units

Answer 1

The correct option is A $\frac{\pi }{6}-\frac{\sqrt{3}}{2}$ sq.units

The circles intersect at $x=0.5$
Hence the required area will be
$=2{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}-\sqrt{1-(x-1{)}^2}.dx$
$=2\times \frac{1}{2}{\left[{\sin }^{-1}\right(x)+x\sqrt{1-x^2}-{\sin }^{-1}(x-1)-(x-1\left)\sqrt{1-(x-1{)}^2}\right]}_{\frac{1}{2}}^1$

$={\left[{\sin }^{-1}\right(x)+x\sqrt{1-x^2}-{\sin }^{-1}(x-1)-(x-1\left)\sqrt{1-(x-1{)}^2}\right]}_{\frac{1}{2}}^1$
$=\frac{\pi }{2}-(\frac{\pi }{6}+\frac{\sqrt{3}}{4}+\frac{\pi }{6}+\frac{\sqrt{3}}{4})$
$=\frac{\pi }{2}-\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
$=\frac{\pi }{6}-\frac{\sqrt{3}}{2}$ sq units.