Area of intersection between two circles [duplicate]

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?

• A similar question exists on Math.SE AFAIR – hjpotter92 May 26 '13 at 12:51
• @hjpotter92 : link to the similar problem please! – jimjim Sep 21 '13 at 3:05
• @Arjang There ^ – hjpotter92 Sep 21 '13 at 7:18
• How much the equation gets more complicated if the circles do NOT pass through the other's center, but intersect each other? – Timo Kähkönen Feb 16 '15 at 8:52

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 $$r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right).$$
• I only corrected the spelling and and I added $$to 2pi/3 in the last line. – Mhenni Benghorbal Jul 13 '13 at 11:28 • ah ok, in that case my comment is directed towards whoever's edit you improved. – Dan Rust Jul 13 '13 at 11:31 We can build an Equilateral triangle between the points, whose side length is r: picture source So we know that the points of intersection are \sqrt{3}r apart, and the angle at them is 60^\circ, by building a rhombus between the dots and centers we know that the angle that "opens" the area is 120^\circ: We can now calculate half the area in question as a circular segment:$$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)$$S is the total area in question • why do we get a different result with lens formula ? is it not a case of symmetric lens ? – George carlin Aug 23 '19 at 14:03 I am taking advantage of the symmetry of the problem We can find the answer by using this formula 4\left(\dfrac{πr^2}{6} - \dfrac{\sqrt{3}}{4r^2}\right) + 2 \dfrac{\sqrt{3}}{4r^2} derivation of this formula is this- Just like triangle ABC we construct triangle CDB. We know that triangle ABC is an equilateral triangle because they are radii of same circle. Therefore angle ABC = 60 degree. Hence we can find area of between chord AB and BC by multiplying the area of a circle with 1/6 i.e.πr^2/6 (because 60 degree/360 degree=1/6) We can subtract the area of triangle ( √3/4 * side^2) from it to find the area of the curved part. In all there are four such congruent parts. Then we can again add the area of the triangle (twice) Hence we will get an answer. • While this might be correct, just writing down a formula is not the best way to answer a question. – egreg Mar 30 '14 at 9:32 • @egreg thanks for editing – steve jobs Mar 30 '14 at 15:28 • Sorry for not giving the derivation.FIRST TIMER :P – steve jobs Mar 30 '14 at 15:58 • Welcome to Mathematics Stack Exchange! Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. – Dan Rust Mar 30 '14 at 16:13 • @DanielRust Thank you so much for your kind words. Will follow all your advices. Thanks again :) – steve jobs Mar 30 '14 at 16:54 One approach is to set up a Boolean function f as$$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.$$Here's how to do this computation in Mathematica. Integrate[ Boole[x^2 + y^2 < r^2 && (x - r)^2 + y^2 < r^2], {x, 0, r}, {y, -r, r}, Assumptions -> r > 0 ] (* Out: -((3*Sqrt[3] - 4*Pi)*r^2)/6 *)  The angle between points of intersection is 2 \pi / 3, so it is:$$ 2 r^2 \cdot \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) 
• The points of intersection are $\sqrt{3}r$ apart, not $\frac{2\pi}{3}$. – SIMEL 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. – Dan Rust May 26 '13 at 13:57