Find the area bounded by the curve `(x-1)^2+y^2=1" and "x^2+y^2=1`.

Explanation

given curve is `x^2+y^2=1` so center `(0,0)` and radius =1
another curve is `(x-1)^2+y^2=1` so center(1,0) and radius =1
from equation (i) and (ii)
`x^2+y^2=1 -> y^2= 1-x^2`
`(x-1)^2 + (1-x^2) = 1`
`x^2+1-2x+1-x^2 =1`
`x=1/2`
if `x=1/2` , then `y=+ - sqrt(3)/2`
hence the intersection point are `(1/(2) , sqrt(3)/(2) ) , (1/(2) , -sqrt(3)/(2) )`
the curve `y= sqrt( 1-(x-1)^2 )` moves from 0 to `1/2`
`y= sqrt(1-x^2)` moves from `1/2` to 1
Required area `= [2 int_0^(1/2)sqrt( 1-(x-1)^2 )dx +int_(1/2)^1sqrt(1-x^2)dx]`
`2[ 1/(2) (x-1)(sqrt( 1-(x-1)^2) +1/(2) sin^-1(x-1)]_0^(1/2) + [ x/(2) sqrt(1-x^2)+ 1/(2) sin^-1(x) ]_(1/2)^1`
on applying the limits ,we get
Area `= [-sqrt(3)/(4) - pi/(6)+ pi/(2) +pi/(2) -sqrt(3)/(4)-pi/(6)]`
`A=[(2pi)/(3) -sqrt(3)/2]` units