Find the area?

Question

Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?

Answer 1

Ooops, I was mistaken at first. To find half of the intersection you need to subtract the area of two 30-60-90 triangles whose hypotenuse is the radius of the circle from the area of 1/3 of the circle. So the entire area of the intersection is:

2/3 of the circle - 4 30-60-90 triangles whose hypotenuse is the radius

=

2pi/3 - sqrt(3)/2

which is approximately 2.094 - .866 = 1.228 square units

Answer 2

Write the equation of the two circles and solve for x and y. Let one circle have center at the origin and the other at (1,0). Then we have.

x² + y² = 1
(x - 1)² + y² = 1

Subtracting we get

2x - 1 = 0
2x = 1
x = 1/2

y² = 1 - x² = 1 - (1/2)² = 3/4
y = √3/2

Find the intercepted arc.

tan θ = (√3/2)/(1/2) = √3
θ = 60°
The intercepted arc is 2θ.
2θ = 120°

The chord length is 2y = √3.

Half of the area of overlap is found as follows.

The intercepted arc is 120°, so the area of the wedge is 1/3 of a circle.

Area wedge = (120/360)π = π/3

Subtract area triangle. It has
base = √3
height = 1/2

Area triangle = (1/2)(1/2)(√3) = √3/4

Half the area of overlap is:

Area wedge - Area triangle = (π/3 - √3/4)

Full area of overlap is:

2(π/3 - √3/4) = 2π/3 - √3/2 ≈ 1.2283697

Answer 3

First make segments from the intersection points to the center of the circles, and a segment connecting the centers. This will create two equilateral triangles with side length of 1 unit.

The area of one of the triangles is 1/2*1*sqrt(3)/2 = sqrt(3)/4
So the combined area of the triangles is sqrt(3)/2

Next your left with 4 circle segments. The area of a segment of a circle = 1/2 * r^2 ((π*Θ/180) - sin Θ). Therefore the area of one segment =

1/2 * 1( (60π/180) - sin 60 )
1/2(π/3 - sqrt(3)/2) (about 0.0906)

to get all four segments multiply that by four =
2(π/3 - sqrt(3)/2) (about 0.362)

Now the total area of the interior is the area of the two triangles and the 4 segments:

= sqrt(3)/2 + 2(1/3π - sqrt(3)/2)
= 2π/3 - sqrt(3)/2

which is 1.228 (rounded to 3 decimals)

Answer 4

The overlap is twice the area of the sector formed by the chord between the intersections of the circle and one arc.
This is the area of the segment formed less the area of the isosceles triangle formed. The angle subtended is 120° or 2π/3 radians. the area of the segment is
(1/2)(2π/3)(1^2)
The area of the triangle is (1sin60°)(1cos60°) = (1/4)√3
The area of one sector, then, is
π/3 - (1/4)√3, and the area of the overlap is
2(π/3 - (1/4)√3) = 1.2284 square units

Answer 5

Let's say circles A and B overlap, and intersect at points C and D. The area of half the overlap would be the area of triangle ABC plus the segments bounded by segment AC and arc AC, and segment BC and arc BC.

The area of triangle ABC is 1/2bh, which is 1/2 * 1 * sqrt(3)/2, or sqrt(3)/4. The area of a segment is the area of sector BAC (or sector ABC) minus the area of triangle ABC. Because triangle ABC is equilateral, angle BAC is exactly 60 degrees, and the area of the sector is exactly 1/6 the area of a complete circle, making it pi/6. Therefore, the area of a segment is pi/6-sqrt(3)/4, or (2pi-3sqrt(3))/12.

The area of half the overlap is the triangle plus two segments, which is sqrt(3)/4 + (2pi-3sqrt(3))/6. Simply double this to get the area of the entire overlap, which is sqrt(3)/2 + (2pi-3sqrt(3))/3, which simplifies to (4pi-3sqrt(3))/6, or 1.228 to three decimal places.

Answer 6

I'll solve this by coordinate geo.
let first circle be at the origin n second be at (1,0)
eqn of first is x^2+y^2=1
n of second is (x-1)^2+y^2=1
their intersection will be at
x^2-(x-1)^2=0
(x-x+1)(x+x-1)=0
x=0.5
for first circle area of quarter section of the overlap will be
integration of sqrt(1-x^2) wrt dx with limits 0 to 0.5
find this integration and multiply the result by 4 (as we have area of a quarter only), this will be your final answer

Answer 7

the overlap is an ellipse
Area of ellipse A = PI ab (see link below for details and figure to locate a and b)
http://www.math.ucsd.edu/~wgarner/math10b/area_ellipse.htm

where b=0.5
and a is : a^2= r^2 - b^2 = 1 - 0.5^2 = 0.75 => a=0.866

so A=PI (0.866)(0.5)=1.36 units