I want to compute the area of a set bounded by three circles. Let $x,y \in [0,1]$, and $\theta \in [0,\pi]$, and consider the three disks $D_1, D_2, D_3$ with unit radius, and centers at the points $(0,0), (x,0), (y \cos \theta, y \sin \theta)$, respectively. What is $A_\theta(x,y) = area(D_1 \cap D_2^c \cap D_3^c)$? (The set $D_1 \cap D_2^c \cap D_3^c = D_1 \setminus (D_2 \cup D_3)$ is an intersection of two crescent shapes.)

It probably isn't possible to give a 'nice' formula for the area in terms of $x,y,$ and $\theta$: all I really care about is a linear approximation for $A_\theta(x,y)$ as a function of $x$ and $y$ as $x,y \to 0$, for fixed $\theta$. I suspect it should have the form

($\star$) $A_\theta(x,y) = c_1(\theta) x + c_2(\theta) y + O(xy, x^2, y^2).$

and I would like to know $c_1$ and $c_2$. Even if $c_1$ and $c_2$ can't be computed, I would still like to know if the Taylor series does have the form $(\star)$, or if not, determine what form it should have.

As an example, one can easily compute the area of a single 'crescent' $D_1 \cap D_2^c$: it essentially comes down to computing the area of intersection of two circles whose radii are $x$ apart. (The set $D_1 \cap D_2^c$ is the crescent shaped region outside the unit disk $D_1$ and inside the shifted disk $D_2$.) This yields

$area(D_1 \cap D_2^c) = \pi-2\arccos(x/2)+\frac{x}{2}\sqrt{4-x^2} = 2x - \frac{x^3}{12} + O(x^5)$ as $x \to 0$.

I can only handle the two trivial cases right now: when $\theta = \pi$, some simple estimates yield the bound $A_\pi(x,y) \leq 4 \sqrt{2} \max(x,y)^3$, so in that case $c_1 = c_2 = 0$, and $A$ decays faster than linearly as $x,y \to 0$. When $\theta = 0$, the intersection reduces to the case above (with just 2 circles), and the area is just $A_0(x,y) = 2 \min(x,y) + O(\min(x,y)^3)$.

I know there are formulas out there for intersections of multiple circles, but so far I've found them too general to apply here. Understanding the set analytically just seems too difficult right now: I can't even get a reasonable expression for the intersection point of the two circles $D_2$ and $D_3$ in terms of $x,y,$ and $\theta$. I think there may be hope in somehow dividing the intersection into two parts: one bounded by $D_1$ and $D_2$, and the other bounded by $D_1$ and $D_3$, and analyzing these separately. But I haven't got that to work yet.

Edit: Added a picture for clarity. The shaded region is what I'm interested in.


  • $\begingroup$ Probably it is my lack of knowledge, but I lost you already when you define $A_\theta$. What does $D_2^2$ and $D_3^c$ mean? $\endgroup$ – EdG Nov 21 '17 at 14:20
  • $\begingroup$ Whoops, just a typo (now fixed): that should be $D_2^c$. $A_\theta$ is just the area of the set $D_1 \setminus (D_2 \cup D_3)$, as a function of $x,y,$ and $\theta$. $\endgroup$ – J Richey Nov 21 '17 at 20:38
  • $\begingroup$ Oh thanks! The picture is very helpful. I will think about it and come back to you if I've an idea :) $\endgroup$ – EdG Nov 22 '17 at 6:14
  • $\begingroup$ BTW, I am sure that in your linear form, $c_1(\theta)$ and $c_2(\theta)$ will be zero, because you linearize around $x=0$ and $y=0$. This is easy to see: if $y$ stays zero, the area will always be zero, regardless of $x$. Vice versa, if $x$ stays zero, the area will always be zero, regardless of $y$. For an analytical expression, this paper might help: dspace.dsto.defence.gov.au/dspace/bitstream/1947/4551/4/… $\endgroup$ – EdG Nov 22 '17 at 7:00

I found an analytical expression for $A_\theta(x, y)$. However, I could not get an easy expression. Define the following areas (for brevity, I omit the dependencies of $x$, $y$ and $\theta$): \begin{align} B &= D_1 \cap D_2^c \\ C &= D_1 \cap D_3 \\ D &= D_1 \cap D_2 \cap D_3 \end{align} Then we have $A=B+D-C$. $B$ and $C$ are easy as you already provided an expression for $B$ and $C$ is almost similar: \begin{align} B &= \pi - 2 \arccos\left(\frac{x}{2}\right)+\frac{x}{2}\sqrt{4-x^2}, \\ C &= 2\arccos\left(\frac{y}{2}\right)-\frac{y}{2}\sqrt{4-y^2}. \end{align} For computing D, I used a paper as a reference. I tried to use the same symbols as much as possible. Let $(x_{ij},y_{ij})$ be the coordinate of the intersection between $D_i$ and $D_j$. Let $d_{23}$ be the distance between the centers of $D_2$ and $D_3$. Let $c_k$ with $k=1,2,3$ be one of the chords of D. Finally, let $\phi$ be the angle between the line going through the centers of $D_2$ and $D_3$ and the horizontal axis (the horizontal axis goes through $D_1$ and $D_2$. Now you can compute $D$ in the following way: \begin{align} x_{12} =& \frac{x}{2}, \\ y_{12} =& \frac12 \sqrt{4-x^2}, \\ x_{13} =& \frac12 \cos(\theta)y + \frac12 \sin(\theta)\sqrt{4-y^2}, \\ y_{13} =& \frac12 \sin(\theta)y - \frac12 \cos(\theta)\sqrt{4-y^2}, \\ d_{23} =& \sqrt{x^2 + y^2 - 2xy\cos(\theta)}, \\ \sin(\phi) =& \frac{y}{d_{23}} \sin(\theta), \\ \cos(\phi) =& \frac{d_{23}^2+x^2-y^2}{2d_{23}x}, \\ x_{23} =& x - \frac12\cos(\phi)d_{23} - \frac12\sin(\phi)\sqrt{4-d_{23}^2}, \\ y_{23} =& \frac12\sin(\phi)d_{23} - \frac12\cos(\phi)\sqrt{4-d_{23}^2}, \\ c_1 =& \sqrt{(x_{12} - x_{13})^2 + (y_{12} - y_{13})^2}, \\ c_2 =& \sqrt{(x_{12} - x_{23})^2 + (y_{12} - y_{23})^2}, \\ c_3 =& \sqrt{(x_{13} - x_{23})^2 + (y_{13} - y_{23})^2}, \\ D =& \frac14 \sqrt{(c_1+c_2+c_3)(c_2+c_3-c_1)(c_1+c_3-c_2)(c_1+c_2-c_3)} + \\ & \,\,\,\,\, \sum_{k=1}^3 \left(\arcsin\left(\frac{c_k}{2}\right) - \frac{c_k}{4}\sqrt{4-c_k^2} \right). \end{align}

I checked the formulas by comparing the result with a numerical estimation (through Monte Carlo) and it seemed right. However, this is a far-from-easy expression. Perhaps one can simplify things. By taking the derivative with respect to $\theta$, you can compute the coefficients of your linear expression (you called them $c_1$ and $c_2$, but they should not be mixed with the chord length $c_k$).

| cite | improve this answer | |
  • $\begingroup$ Oh gosh. I guess I could try plugging this into Mathematica. I saw this paper, but was too afraid to write down the formulas: thanks for doing that in a clear way! $\endgroup$ – J Richey Nov 27 '17 at 20:12
  • $\begingroup$ Why does $A = B + D - C$ hold? This seems incorrect to me. $\endgroup$ – J Richey Dec 5 '17 at 21:53
  • $\begingroup$ I can see the following (trivial) equality: if $B = D_1 \cap D_2, C = D_1 \cap D_3$, and $D = D_1 \cap D_2 \cap D_3$, then $A = D_1 - B - C + D$. $\endgroup$ – J Richey Dec 5 '17 at 22:01
  • $\begingroup$ Oops, I wrote $\cup$ instead of $\cap$... I changed it. Is it now clear for you? $\endgroup$ – EdG Dec 6 '17 at 3:19
  • $\begingroup$ BTW, the equality that you propose would also work :) $\endgroup$ – EdG Dec 6 '17 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.