# Area of intersection between two circles

Suppose you have 2 circles that intersect each other in such a way that each circle passes through the other's center. What is the area between the circle(or common area) i.e. area between the centres of the circles?

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A similar question exists on Math.SE AFAIR –  hjpotter92 May 26 '13 at 12:51
@hjpotter92 : link to the similar problem please! –  Arjang Sep 21 '13 at 3:05
possible duplicate of Expected area of the intersection of two circles –  hjpotter92 Sep 21 '13 at 7:18
@Arjang There ^ –  hjpotter92 Sep 21 '13 at 7:18

We can build an Equilateral triangle between the points, whose side length is $r$:

picture source

So we know that the points of intersection are $\sqrt{3}r$ apart, and the angle at them is $60^\circ$, by building a rhombus between the dots and centers we know that the angle that "opens" the area is $120^\circ$:

We can now calculate half the area in question as a circular segment:

$$S=2\left[\frac{r^2}{2}\left(\frac{2\pi}{3}-\sin\left(\frac{2\pi}{3}\right)\right)\right]=r^2\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)$$

$S$ is the total area in question

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Label the center of the first circle $C$ and the center of the second circle $C'$. Label one of the points of intersection of the two circles $A$ and the other $B$. Let the radius of the circles be $r>0$. It should be clear that the following lengths are all equal to $r$. $AC$, $AC'$, $BC$, $BC'$, $CC'$. With a simple application of Pythagoras' Theorem, we get that the length of the line segment $AB$ is $\sqrt{3}r$.

With some basic trigonometry, we find the angles $\angle ACB=\angle AC'B=\dfrac{2\pi}{3}$. So, the area of one half of the intersection is the area of a circular segment with angle $\theta=\dfrac{2\pi}{3}$ and radius $r$, which gives an area of $\dfrac{r^2}{2}(\theta-\sin\theta)=\dfrac{r^2}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)$ and so the area of the entire intersection is twice this. This gives an area of $$r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right).$$

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@MhenniBenghorbal I'm a little confused by the edit. The problem states that both circles intersect and also pass through each others centers. This happens if and only if they have the same radius. –  Daniel Rust Jul 13 '13 at 11:24
I only corrected the spelling and and I added $$to 2pi/3 in the last line. – Mhenni Benghorbal Jul 13 '13 at 11:28 ah ok, in that case my comment is directed towards whoever's edit you improved. – Daniel Rust Jul 13 '13 at 11:31 add comment The angle between points of intersection is 2 \pi / 3, so it is:$$ 2 r^2 \cdot \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) $$- explanation? and how did you get POI distances as 2pi/3 @vonbrand – Adeetya May 26 '13 at 13:03 @Adeetya, consider how a hexagon is constructed. – vonbrand May 26 '13 at 13:05 The points of intersection are \sqrt{3}r apart, not \frac{2\pi}{3}. – Ilya Melamed May 26 '13 at 13:32 I believe vonbrand meant the angle between the points with respect to the center of the circles. There does seem to be a mistake with the his expression though, as the area should grow linearly with r^2, not r^4 as his does. – Daniel Rust May 26 '13 at 13:57 @DanielRust, you are right. Last minute factorization went wrong. Thanks! – vonbrand May 26 '13 at 23:02 add comment One approach is to set up a Boolean function f as$$f(x,y) = \left\{ \begin{array}{ll} 1 & \text{if } x^2+y^2<r^2 \text{ and } (x-r)^2+y^2<r^2 \\ 0 & \text{else}. \end{array}\right. $$The area can then be expressed as$$\int_0^r \int_{-r}^r f(x,y) \, dx \, dy.

Here's how to do this computation in Mathematica.

Integrate[
Boole[x^2 + y^2 < r^2 && (x - r)^2 + y^2 < r^2],
{x, 0, r}, {y, -r, r},
Assumptions -> r > 0
]

(* Out: -((3*Sqrt[3] - 4*Pi)*r^2)/6 *)

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