what is the area covered by both the same-radius circles who go through each other's centre?

Question

2 circles of same radius, partially overlap. each go through other's centre. need to know the area(shaped like an eye) covered by both of them.

Answer 1

Let M, N be the centers of the circles. Let A, B the points of intersection. Let P be a point on the circle about M between A and B, and Q a similar point on the other circle.

It is well-known that the arcs APB and AQB have measure 60 degrees.

Split the overlap area in two halves by drawing line AB. Each half is a circle segment. We will calculate the area of segment APB.

It is the difference of areas of sector MAPB and triangle MAB.

Sector MAPB is 1/6 of the circle's area:
[1] ... 1/6 * pi * R^2

Triangle MAPB is equilateral triangle, with base R and height 1/2 * sqrt(3) * R:
[2] ... 1/4 * sqrt(3) * R^2

The difference is area of segment APB:
[3] ... [1/6 * pi - 1/4 * sqrt(3)] * R^2

The overlap consists of two of these segments:
[4] ... [1/3 pi - 1/2 sqrt 3] R^2

This is 0.1812 R^2, or 5.8% of the area of a single circle.

Answer 2

The angle of the 'eye' is 60 degrees. If you look at one-half of the eye, you will see it is the difference of this angular sector and an equilateral triangle. The sector has area equal to (1/6) of the circle. The triangle has area equal to (sqrt(3)/4) r^2. So one-half of the eye has area
[pi/6 -sqrt(3)/4]r^2.
Double this to get the area of the whole eye:
[pi/3-sqrt(3)/2]r^2.

Answer 3

( (2/3)*pi - sqrt(3)/2 ) * r^2
approx. 1.228369699 * r^2

Two equilateral triangles which are (1/4)*sqrt(3)*r^2 each plus
four segments (area bounded by a chord and an arc) which are (1/6)*pi*r^2 - (1/4)*sqrt(3)*r^2 each

so:

(2/4)*sqrt(3)*r^2 + (4/6)*pi*r^2 - (4/4)*sqrt(3)*r^2
(1/2)*sqrt(3)*r^2 + (2/3)*pi*r^2 - sqrt(3)*r^2
(1/2) * ( (4/3)*pi*r^2 - sqrt(3)*r^2 )
(2/3)*pi*r^2 - (1/2)*sqrt(3)*r^2
=
( (2/3)*pi - sqrt(3)/2 ) * r^2


edit: My first answer was wrong. I had to change it.

It should be a little more than one third the area of the circle or a little more than 1.047197551 * r^2 so my solution checks out!

To the other three people who have answered so far... how can it possibly be less than one third the area of one of the circles?

Answer 4

Try this formula:

Shaped Area (like an eye) = (134R^3 - 32R²) / (192R - 48)
R = Radius of the circles.
The total area works out to be about 22∙3% of one of the single circles.