Question

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

Answer 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 = ±\(\frac{\sqrt3}{2}\)

So the two points, A and B where the circles (1) and (2) meet are A = \(\Big(\frac{1}{2},\frac{\sqrt3}{2}\Big)\) and B = \(\Big(\frac{1}{2},\frac{\sqrt3}{2}\Big)\)

The line connecting AB, will be intersecting the x-axis at D = \(\Big(\frac{1}{2},0\Big)\)

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

y² = 1 - x²

y = \(\sqrt{(1-x^2)}\) ....(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 y = \(\sqrt{(1-x^2)}\) is the equation.

The Area of the required region = \(\Big(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\Big)\)sq.units