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