Let $C_1$ and $C_2$ be two circles of radii $r_1$ and $r_2$ respectively whose centers are at a distance $d$ from each other. Assume, without loss of generality, that $r_1 \geq r_2$. What is the intersection area of these two circles?

If $d \geq r_1 + r_2$, the circles intersect at most up to a point (when $d = r_1 + r_2$) and therefore the intersection area is zero. On the other extreme, if $d + r_2 \leq r_1$, circle $C_2$ is entirely contained within $C_1$ and the intersection area is the area of $C_2$ itself: $\pi r_2^2$. The challenging case happens when both $d \lt r_1 + r_2$ and $d + r_2 \gt r_1$ are satisfied, i.e., when the the circles intersect only partially but the intersection area is more than simply a point. Rearranging the second inequality, we obtain $r_1 - r_2 \lt d \lt r_1 + r_2$, so we will assume this to be the case from now on.

To solve this problem, we will make use of a Cartesian coordinate system with origin at the center of circle $C_1$ such that the center of $C_2$ is at $(d,0)$ as shown on figure 1.

Fig. 1: | Two intersecting circles $C_1$ (blue) and $C_2$ (red) of radii $r_1$ and $r_2$ respectively. The distance between the centers of the circles is $d = d_1 + d_2$, where $d_1$ is the $x$ coordinate of the intersection points and $d_2 = d - d_1$. Notice that $d_1 \geq 0$ since these points are always located to the right of the center of $C_1$, but $d_2$ may be negative when $r_2 \lt r_1$ since, in this case, the intersection points will eventually fall to the right of the center of $C_2$ as we move $C_2$ to the left, making $d \lt d_1$ and therefore $d_2 \lt 0$. |

The circles $C_1$ and $C_2$ are described by the following equations respectively: $$ \begin{eqnarray} x^2 + y^2 &=& r_1^2 \label{post_8d6ca3d82151bad815f78addf9b5c1c6_c1}\\[5pt] (x - d)^2 + y^2 &=& r_2^2 \\[5pt] \end{eqnarray} $$ At the intersection points, we have $x = d_1$. To determine $d_1$, we can replace $x$ with $d_1$ and isolate $y^2$ on both equations above to get: $$ r_1^2 - d_1^2 = r_2^2 - (d_1 - d)^2 $$ Solving for $d_1$ is a simple task: $$ r_1^2 - d_1^2 = r_2^2 - d_1^2 + 2d_1d - d^2 \Longrightarrow d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1} $$ From equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}, we can see that $d_1 \geq 0$ since $r_1 \geq r_2$. The intersection area is the sum of the blue and red areas shown on figure 1, which we refer to as $A_1$ and $A_2$ respectively. We then have that: $$ \begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \label{%INDEX_eq_A1_def} \\[5pt] A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \end{eqnarray} $$ where the factors of $2$ come from the fact that each integral above accounts for only half of the area of the associated region (only points on and above the $x$ axis are taken into account); the results must then be multiplied by two so that the areas below the $x$ axis are taken into account as well.

The computation of these integrals is straightforward. Before we proceed, notice first that: $$ \begin{eqnarray} A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \nonumber \\[5pt] &=& 2\int_{- r_2}^{d_1 - d} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d - d_1}^{r_2} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d_2}^{r_2} \sqrt{r_2^2 - x^2}dx \label{%INDEX_eq_A2} \end{eqnarray} $$ where above we used the fact that $d_2 = d - d_1$. This is the same as equation \eqref{%INDEX_eq_A1_def} if we apply the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$. Therefore, by computing $A_1$, we will immediately obtain $A_2$ as well. Let's then compute $A_1$ first: $$ \begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \nonumber\\[5pt] &=& 2r_1 \int_{d_1}^{r_1} \sqrt{1 - \left(\frac{x}{r_1}\right)^2}dx \nonumber\\[5pt] &=& 2r_1^2 \int_{d_1/r_1}^{1} \sqrt{1 - x^2}dx \label{%INDEX_eq_A1} \end{eqnarray} $$ All we need to do now is to integrate $\sqrt{1 - x^2}$. The process is straightforward if we use integration by parts: $$ \begin{eqnarray} \int \sqrt{1 - x^2}dx &=& x \sqrt{1 - x^2} - \int x \left(\frac{-x}{\sqrt{1 - x^2}}\right) dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} + \int \frac{x^2 - 1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} - \int \sqrt{1 - x^2} dx + \sin^{-1}(x) \end{eqnarray} $$ Therefore: $$ \int \sqrt{1 - x^2}dx = \frac{1}{2}\left( x \sqrt{1 - x^2} + \sin^{-1}(x) \right) \label{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} $$ Using equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} on equation \eqref{%INDEX_eq_A1} yields: $$ \begin{eqnarray} A_1 &=& r_1^2 \left( \frac{\pi}{2} - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} - \sin^{-1}\left(\frac{d_1}{r_1}\right) \right) \nonumber\\[5pt] &=& r_1^2 \left( \cos^{-1}\left(\frac{d_1}{r_1}\right) - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} \right) \nonumber\\[5pt] &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1 \sqrt{r_1^2 - d_1^2} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final} \end{eqnarray} $$ where above we used the fact that $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$ for any $\alpha$ in $[-1,1]$. This fact is easy to prove: $$ \cos\left(\frac{\pi}{2} - \sin^{-1}(\alpha)\right) = \cos\left(\frac{\pi}{2}\right)\cos(\sin^{-1}(\alpha)) + \sin\left(\frac{\pi}{2}\right)\sin(\sin^{-1}(\alpha)) = \alpha $$ and therefore $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$. As discussed above, we can now obtain $A_2$ directly by doing the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$ on the expression for $A_1$ on equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final}: $$ A_2 = r_2^2 \cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2 \sqrt{r_2^2 - d_2^2} $$ The sum of $A_1$ and $A_2$ is the intersection area of the circles: $$ \boxed{ \begin{eqnarray} A_{\textrm{intersection}} &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1\sqrt{r_1^2 - d_1^2} \nonumber \\[5pt] &+& r_2^2\cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2\sqrt{r_2^2 - d_2^2} \nonumber \end{eqnarray} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} $$ where: $$ \boxed{ d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} } \quad \textrm{ and } \quad \boxed{ d_2 = d - d_1 = \displaystyle\frac{r_2^2 - r_1^2 + d^2}{2d} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1_final} $$

### Summary

Given two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ respectively (with $r_1 \geq r_2$) whose center points are at a distance $d$ from each other, the intersection area of the circles is:

1. | zero, if $d \geq r_1 + r_2$, since in this case the circles intersect at most up to a point. |

2. | $\pi r_2^2$, if $d \leq r_1 - r_2$, since in this case $C_2$ is entirely contained within $C_1$. |

3. | given by equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} in all other cases. |

## Comments

Best regards

L.

Thanks in advance.

If the article is hard to follow, I suggest you read only the summary at the end as it tells you how to compute the intersection area in every possible scenario.

Please notice that the derivation above did not assume that the center of $C_2$ was either inside or outside of $C_1$, so it works in both situations.

Notice that there is no loss of generality in this assumption because the particular label assigned to each circle plays no role in the derivation.

It is actually possible that equations (14) and (15) are valid even when $r_1 \leq r_2$, but this is not an obvious fact on the derivation above.

Thanks in advance!

Stefan

$\displaystyle \int_{x \,=\, d_1}^{x \,=\, r_1} \sqrt{1 - \left(\frac{x}{r_1}\right)^2}dx

= \int_{u \, = \, d_1/r_1}^{u \, = \, 1} \sqrt{1 - u^2}\,r_1 du

$

After that, I simply replaced $u$ with $x$ again.

I also enjoyed reading your concisely written and well designed post. Would you, by chance, know if there's a formula/algorithm for determining the intersection area when you have only slices/wedges of two circles?

I would like to use your derivation to model the spherically symmetrical expansion of an eruption cloud on the surface of a spherical comet nucleus. In this case the source of the eruption is at the centre of C2, which lies inside C1 (hence d<r1). After the eruption, the value of r2 increases linearly with time. You say your derivation is valid up until r2 equals r1, but what about the condition where r2 > r1? Do the two circles interchange so that C2 becomes C1 and vice versa and so a solution is possible?

As I mentioned on a previous comment, it is possible that equations (14) and (15) are valid even when $r_1 \lt r_2$, but the derivation above does not make it obvious whether that is true or not.

I was not sure about your earlier comment - thanks for your confirmation. I shall go ahead and model the total reflectance of the nucleus and an expanding dust cloud of different reflectivity. We have observations from 2019 of such an eruption on a comet and comparing the model with the data we have should set constraints on the location of the eruption relative to the centre of the nucleus. Of course, I shall acknowledge your contribution in the paper.

Richard

Do you think it would be possible to approach this problem in a similar manner to how you did above by considering an n-gon approximation for each circle? I'm working on a problem where I have to integrate a function f(x,y) over the intersection of two circles. I'm trying to avoid the radical signs.

I was wondering how you would go about this when the smaller circle also has some y component to it (i.e. the centerpoints are not directly horizontal to one another).

The details are unfortunately too complicated to be explained here, but feel free to reach me via e-mail (diego@assencio.com) if you wish to get additional pointers.

The results above are therefore correct in all situations, provided that you define $C_1$ and $C_2$ such that $r_1 \geq r_2$.

@Joseph : Are you dealing with FSO laser beam misalignment problems?

Diego is right about the setup as per the objective. However, if you are trying to integrate over the intensity of a Gaussian beam which has non-uniform intensity over the large circle, a y-translation for misalignment IS indeed better. My advice is to forget it because it makes life harder than it already is.

Please let me know the source of this post so that I can refer to it in my article.

Thanks a lot