Find the area of the region enclosed between the circles x2 +y2=1 and (x-1)2+y2=1

Given the boundaries of the area to be found are,


• First circle, x2 + y2 = 1 ---(1)


• Second circle, (x-1)2 + y2 = 1 ---- (2)


From the equation, of the first circle, x2 + y2 = 1


• the vertex at (0,0) i.e. the origin


• the radius is 1 unit.


From the equation, of the second circle, (x-1)2 + y2 = 1


• the vertex at (1,0) i.e. the origin


• the radius is 1 unit.


Now to find the point of intersection of (1) and (2), substitute y2 = 1-x2 in (2)


(x-1)2 + (1-x2) = 1


x2 + 1 – 2x +1-x2 = 1



Substituting x in (1), we get


So the two points, A and B where the circles (1) and (2) meet are and


The line connecting AB, will be intersecting the x-axis at


As x and y have even powers for both the circles, they will be symmetrical about the x-axis and y-axis.


Here the circle, x2 + y2 = 1, can be re-written as



----- (3)



Now, the area to be found will be the area is


Area of the required region = Area of OABC.


Area of OABC = Area of AOC + Area of BOC


[area of AOC = area of BOC as the circles are symmetrical about the y-axis]


Area of OABC = 2 × Area of AOC


Area of OABC = 2 (Area of OAD + Area of ADC)


[area of OAD = area of ADC as the circles are symmetrical about the x-axis]


Area of OABC = 2 (2 × Area of ADC)


Area of OABC = 4 × Area of ADC


Area of ADC is under the first circle, thus is the equation.




[Using the formula, ]




[sin-1(1) = 90° and ]


The Area of the required region sq. units


1