Find the function does describe the the percentage of the area that each circle overlaps

Question

I saw this question, yesterday and it got me thinking, what function does describe the the percentage of the area that each circle overlaps.

In that diagram it is given that the distance between the center of the circles is $1.5r$ but I want to generalize that to $nr$. The circles has the same radius.

Question: what function describes the problem I want to solve?

I tried using the answer that was given in the original problem, but that didn't bring me further. Thanks for any advice.

Answer 1

Well, let's solve this by using some graphics. I define the following diagram:

The equation of a circle is given by:

$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$

Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.

We have two circles and they can be described using (which can be seen from the diagram above):

1. $$\left(x-\alpha_1\right)^2+\text{y}^2=\text{R}^2\tag2$$
2. $$\left(x-\alpha_2\right)^2+\text{y}^2=\text{R}^2\tag3$$

It is also not hard to see that $\alpha_1=\frac{\text{R}\Delta}{2}$ and $\alpha_2=-\alpha_1=-\frac{\text{R}\Delta}{2}$, using the diagram above.

Now, we can look at the following diagram:

If we combine both diagrams, we can see that $\epsilon_1=\left|\text{R}-\frac{\text{R}\Delta}{2}\right|=\text{R}\left|1-\frac{\Delta}{2}\right|$ and $\epsilon_2=-\epsilon_1=-\text{R}\left|1-\frac{\Delta}{2}\right|$. And using $(2)$ and/or $(3)$ we can solve for $\lambda_1$ and $\lambda_2$:

$$\left(0-\alpha_1\right)^2+\lambda_1^2=\text{R}^2\space\Longleftrightarrow\space\lambda_1=\sqrt{\left|\text{R}^2-\alpha_1^2\right|}\space\wedge\space\lambda_2=-\lambda_1=-\sqrt{\left|\text{R}^2-\alpha_1^2\right|}\tag4$$

Now, we take a look at the last diagram:

Using the first and the third diagram it is not hard to see that $\varphi_1=\text{R}+\frac{\text{R}\Delta}{2}=\text{R}\left(1+\frac{\Delta}{2}\right)$ and $\varphi_2=-\varphi_1=-\text{R}-\frac{\text{R}\Delta}{2}=-\text{R}\left(1+\frac{\Delta}{2}\right)$.

Now, we can solve $(2)$ and $(3)$ for the equation in the first quadrant in the first diagram:

• $$\left(x-\alpha_1\right)^2+\text{y}^2=\text{R}^2\space\Longrightarrow\space\text{y}=\sqrt{\text{R}^2-\left(x-\alpha_1\right)^2}\tag5$$
• $$\left(x-\alpha_2\right)^2+\text{y}^2=\text{R}^2\space\Longrightarrow\space\text{y}=\sqrt{\text{R}^2-\left(x-\alpha_2\right)^2}\tag6$$

Now, we can integrate to find the areas (using that a circle is symmetric):

• The total area in diagram 3 is given by: $$\mathcal{A}_1:=4\int_0^{\varphi_1}\sqrt{\text{R}^2-\left(x-\alpha_1\right)^2}\space\text{d}x=$$ $$\frac{\text{R}^2}{2}\left(2\pi+\Delta\sqrt{4-\Delta^2}+4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)\right)\tag7$$
• The total area in diagram 2 is given by: $$\mathcal{A}_2:=4\int_0^{\epsilon_1}\sqrt{\text{R}^2-\left(x-\alpha_2\right)^2}\space\text{d}x=$$ $$\frac{\text{R}^2}{2}\left(2\pi-\Delta\sqrt{4-\Delta^2}-4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)\right)\tag8$$

So, the desired fraction is given by:

$$\eta:=\frac{\mathcal{A}_2}{\mathcal{A}_1}=\frac{\frac{\text{R}^2}{2}\left(2\pi-\Delta\sqrt{4-\Delta^2}-4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)\right)}{\frac{\text{R}^2}{2}\left(2\pi+\Delta\sqrt{4-\Delta^2}+4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)\right)}=$$ $$\frac{2\pi-\Delta\sqrt{4-\Delta^2}-4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)}{2\pi+\Delta\sqrt{4-\Delta^2}+4\arctan\left(\frac{\Delta}{\sqrt{4-\Delta^2}}\right)}\tag9$$

Note: these calculations only make sense when $0\le\Delta\le2$ and when $\Delta\to2$ we get (because we get a undetermined form):

$$\lim_{\Delta\to2^-}\eta=0\tag{10}$$

As the last step, we can plot $\eta$ as a function of $\Delta$:

---

If you use Mathematica, you can use some code of me. In order to plot a complete diagram, you can use:

R =;
\[CapitalDelta] =;
\[Alpha]1 = (R*\[CapitalDelta])/2;
\[Alpha]2 = -\[Alpha]1;
\[Epsilon]1 = R*Abs[1 - (\[CapitalDelta]/2)];
\[Epsilon]2 = -\[Epsilon]1;
\[Lambda]1 = Sqrt[Abs[R^2 - \[Alpha]1^2]];
\[Lambda]2 = -\[Lambda]1;
\[Phi]1 = R*(1 + (\[CapitalDelta]/2));
\[Phi]2 = -\[Phi]1;
ContourPlot[{(x - \[Alpha]1)^2 + y^2 == 
   R^2, (x - \[Alpha]2)^2 + y^2 == R^2}, {x, -2 R, 2 R}, {y, -2 R, 
  2 R}, GridLines -> {{0, \[Alpha]1, \[Alpha]2, \[Epsilon]1, \
\[Epsilon]2, \[Phi]1, \[Phi]2}, {0, \[Lambda]1, \[Lambda]2}}]

In order to evaluate the integral $(7)$, you can use:

FullSimplify[
 Integrate[
  4 Sqrt[R^2 - (x - ((R*\[CapitalDelta])/2))^2], {x, 0, 
   R*(1 + (\[CapitalDelta]/2))}, Assumptions -> R > 0 && \[CapitalDelta] > 0]]

In order to evaluate the integral $(8)$, you can use:

FullSimplify[
 Integrate[
  4 Sqrt[R^2 - (x - (-(R*\[CapitalDelta])/2))^2], {x, 0, 
   R*Abs[1 - (\[CapitalDelta]/2)]}, Assumptions -> R > 0 && \[CapitalDelta] > 0]]

And so $\eta$ is found by:

FullSimplify[(1/
    2 R^2 (2 \[Pi] - \[CapitalDelta] Sqrt[4 - \[CapitalDelta]^2] - 
      4 ArcTan[\[CapitalDelta]/Sqrt[4 - \[CapitalDelta]^2]]))/(1/
    2 R^2 (2 \[Pi] + \[CapitalDelta] Sqrt[4 - \[CapitalDelta]^2] + 
      4 ArcTan[\[CapitalDelta]/Sqrt[4 - \[CapitalDelta]^2]])), 
 Assumptions -> R > 0 && \[CapitalDelta] > 0]

And the plot of $\eta$ as a function of $\Delta$ is made using:

Plot[(2 \[Pi] - \[CapitalDelta] Sqrt[4 - \[CapitalDelta]^2] - 
  4 ArcTan[\[CapitalDelta]/Sqrt[4 - \[CapitalDelta]^2]])/(
 2 \[Pi] + \[CapitalDelta] Sqrt[4 - \[CapitalDelta]^2] + 
  4 ArcTan[\[CapitalDelta]/Sqrt[4 - \[CapitalDelta]^2]]), {\[CapitalDelta], 0, 2}, 
 AxesLabel -> {HoldForm[\[CapitalDelta]], HoldForm[\[Eta]]}, 
 PlotLabel -> None, LabelStyle -> {GrayLevel[0]}]