As distance between overlapping circles decreases, find radius so that total area is constant and proportion of overlapping area increases constantly

Question

Take 2 circles of radius $r$ distant (center to center) by $d$. They start by barely touching each other, $d = 2r$.

Then, as the distance decreases, I have 2 requirements: I want

1. the total area to stay constant, and
2. the rate at which the proportion of the overlapping area (relative to the total area) increases, necessary to satisfy 1., to stay constant as well.

I know that:

• the area of the circles is $A_C(r)=2 \pi r^2$
• the overlapping area is given by $A_O(r,d) = 2 r^2 \text{cos}^{-1}\left(\frac{d}{2r}\right) - \frac{1}{2}d\sqrt{4r^2 - d^2}$
• therefore the total area is $A_T(r,d) = A_C(r) - A_O(r,d) = 2 \pi r^2 - 2 r^2 \text{cos}^{-1}\left(\frac{d}{2r}\right) + \frac{1}{2}d\sqrt{4r^2 - d^2}$

Requirement 1 tells me that, for all $r$ and $d$, and calling $r_0$ the initial radius, it must be that $A_T(r,d) = A_C(r_0)$.

Requirement 2 tells me that the change in the proportion of overlap, $\frac{A_O(r,d)}{A_T(r,d)}$, should be constant.

And, well... here I'm stuck.

I've done some testing in JavaScript and I have the feeling that Requirements 1 and 2 cannot be satisfied at the same time, but I'm not quite sure.

Answer 1

Tl;dr: Both conditions are possible, but condition 2 cannot be satisfied in terms of elementary functions.

Here are the equations that satisfy condition 1: $$ \begin{align} r&=\sqrt{\frac{A}{2\arccos\left(-s\right)+2s\sqrt{1-s^{2}}}}\\ d&=2rs \end{align} $$ Where $A$ is the constant area and $s$ is a parameter that you slide from $s=0$ to $s=1$. At $0$, the two circles are completely merged, and at $1$ they are completely seperate.

Messy Desmos graph: https://www.desmos.com/calculator/aeahpotvof $$ \! $$ Here's how I found these:

I started with your equation for total area (written in a slightly different way) $$ A=2\pi r^{2}-2r^{2}\arccos\left(\frac{d}{2r}\right)+\frac{d}{2}\sqrt{4r^{2}-d^{2}} $$ Now, as both $d$ and $r$ appear both inside and outside inverse cosine, we will not be able to separate them. However, I substituted in a new variable, called $s$ $$ \begin{align} s&=\frac d{2r}\\ A&=2\pi r^{2}-2r^{2}\arccos\left(s\right)+\frac{d}{2}\sqrt{4r^{2}-d^{2}}\\ &=2\pi r^{2}-2r^{2}\arccos\left(s\right)+2r^{2}s\sqrt{1-s^{2}}\\ &=2r^2\left(\pi -\arccos\left(s\right)+s\sqrt{1-s^{2}}\right)\\ &=2r^2\left(\arccos(-s)+s\sqrt{1-s^{2}}\right)\\ r&=\sqrt{\frac A{2r^2\left(\arccos(-s)+s\sqrt{1-s^{2}}\right)}} \end{align} $$ Now that we have $r$ in terms of $s$, we can easily rearrange our definition of $s$ to get $d$ in terms of $s$ and $r$, and condition 1 will be fulfilled. $$ \! $$ Here's why it's impossible to fulfill condition 2 in terms of elementary functions:

I will skip the working out, as it is tedius plug-and-chug, but substituting the solutions for $r$ and $s$ into the equation for overlapping area, we get the following: $$ A_O(s)=A\frac{\arccos\left(s\right)-s\sqrt{1-s^{2}}}{\arccos\left(-s\right)+s\sqrt{1-s^{2}}} $$ In order to make the overlap's growth rate constant, we must find its inverse, and then plug its inverse into itself, as $f\left(f^{-1}(x)\right)=x$ is a line. However, my attempts to find an inverse led me to a function that is known to have no inverse in terms of elementary functions.

Step 1 was to use the substitution $s=\cos(x)$. This can be undone later, and it transforms the function into a more manageable form. $$ A_O(x)=\frac{x-\sin\left(x\right)\cos\left(x\right)}{\pi-x+\sin\left(x\right)\cos\left(x\right)} $$ $$ \begin{align} \frac{1}{A_O}&=\frac{\pi-x+\sin\left(x\right)\cos\left(x\right)}{x-\sin\left(x\right)\cos\left(x\right)}\\ \frac{1}{A_O}&=\frac\pi{x-\sin\left(x\right)\cos\left(x\right)}-1\\ \frac{1}{A_O}+1&=\frac\pi{x-\sin\left(x\right)\cos\left(x\right)}\\ \frac\pi{\frac{1}{A_O}+1}&=x-\sin\left(x\right)\cos\left(x\right)\\ \frac{2\pi}{\frac{1}{A_O}+1}&=2x-\sin(2x)\\ \end{align} $$ However, $2x-\sin(2x)$ is just a stretched version of $x-\sin(x)$, which is known to have no inverse. Since we cannot invert it, we cannot get $x$ in terms of $A_O$, and therefore we cannot get $s$ in terms of $A_O$. This means that we cannot find a transformation to put into $s$ that will make the overlapping area linear.