Find the area of the region bounded by the curves `x^(2)+y^(2)=4` and `(x-2)^(2)+y^(2)=4.`

Explanation

`x^2+y^2=4` is a circle with its centre at O (0,0) and radius =2 units and `(x-2)^2+y^2=4` is a circle with its centre at C(2,0) and The given circles are
`x^2y^2=4`
and `(x-2)^2=4` is circle with its centre at C(2,0) and radius =2 units
The given circles are
`x^2+y^2=4`
and `(x-2)^+y^2=4`
Eliminating y form (i) and (ii) we get
`4-x^2=4 -(x-2)^2rArr 4x = 4 rArr x =1`
Putting x=1 in (i) we get `y^2= 3 rArr y = pm sqrt(3)`
Thus the points of intersection of the two circles are A `A(1,sqrt3)` and `B(1,-sqrt(3)`
Both the circles are symmetrical about the x-axis. Required are =2(area AOCA)
=2(area AODA + area CADC)
`=2[underset(0)overset(2)int "y dx for curve (ii)" + underset(2)overset(4)int"y dx for curve"]`
`=2[underset(0)overset(2)intsqrt(6x)dx+ overset(4)underset(2)intsqrt(16-x^2)dx]`
`={[(2sqrt(6))/(3)x^(3//2)]_0^2+[xsqrt((16-x^2)/2)+16/2 sin^(-1)""x/4]_0^4}`
`{((2sqrt(6))/(3).2^(3//2)-0)+8 sin^-1""1-(2sqrt(3)+8sin^(-1)""1/2)}`
`=2((8sqrt(3))/(3)+4 pi-2sqrt3-(4pi)/(3))=2((2sqrt(3))/(3)+(8pi)/(3))`
`=4/3(sqrt(3)+4pi)"sq units "`
Hence , the required area is `4/3(sqrt(3)+pi)` sq units .