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

*

*the total area to stay constant, and

*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.
 A: 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
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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.
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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.
