
Consider
x2 + y2 = 1 …… (1)
(x – 1)2 + y2 = 1 …… (2)
Here the intersection at the point which is obtained by solving both the equations is
y2 = 1 – x2
By substituting it in equation (2) we get
(x – 1)2 + 1 – x2 = 1
On further simplification
(x – 1)2 – x2 = 0
We can write it as
(x – 1 – x) (x – 1 + x) = 0
Here we get
– 2x + 1 = 0 where x = ½
Using this in equation (1) we get y = ± √3/2
So both the equations intersect at point A (1/2, √3/2) and B (1/2, –√3/2)
(0, 0) is the center of first circle and radius 1
Similarly (1, 0) is the center of second circle and radius is 1
Here both the circles are symmetrical about x-axis and the required area is shaded here.
So the required area = area OACB = 2 (area OAC)
= 2 [area of OAD + area DCA]
