find the area of the region enclosed between the two circles x^2+y^2=4 and (x-2)^2+y^2=4

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find the area of the region enclosed between the two circles
x^2+y^2=4 and (x-2)^2+y^2=4

Manbar Singh · Accepted Answer

The given equations are :x2 + y2 = 4 1x - 22 + y2 = 4 2Equation 1
represents a circle having the centre at the origin O0, 0 and
radius 2 units Equation 2 represents a circle having the centre at
O'2, 0 and radius 2 units POINTS OF INTERSECTION :From 1, we have
,y2 = 4 - x2put the value of y2 in 2, we getx - 22 + 4 - x2 =
4⇒x2 + 4 - 4x + 4 - x2 = 4⇒4 - 4x = 0⇒x = 1When x =
1, then y2 = 4 - 1 = 3⇒y = ±3so the points of
intersection of 1 and 2 are : A1, 3 and B1, -3

Required area = area of region AO'BOA

area = 2∫014 - x - 22 dx + 2∫124 - x2 dxso, area = 2 I1 +
2I2 = 2I1 + I2 3now, I1 = ∫014 - x - 22 dx= x-224 - x - 22 +
42sin-1x-2201=-123 + 2sin-1-12 - 0 + 2 sin-1-1=-32 -2 sin-112 + 2
sin-11=-32 - 2 × π6 + 2×π2 =32 - π3 +
πI1=2π3 - 32

I2 = ∫124 - x2 dx= x24 - x2 + 42 sin-1x212=0 + 2sin-11 - 123 +
2 sin-112= 2 × π2 - 32 - 2 × π6= π - π3 -
32I2 = 2π3 - 32

From 3, we get,area = 2I1 + I2area = 22π3 - 32 + 2π3 - 32
required area = 24π3 - 3 square units