How can I formulate the area of this shape so that the functional inverse of the expression can be found? I have been working on this problem for a few weeks now, and I am quite stumped. I will walk you through the goal and why my efforts have failed before finally asking the specific question teased in the title.
First, I need to introduce a square with side lengths $L$ and a circle, with radius $r$, that shares its center with the square.

Given this setup, we might ask "What percent of the square's area is covered by the circle's area?" This could be found as a function of the circle's radius, say $P(r)$. To be perfectly general it would also be a function of $L$, but I'm perfectly okay with assuming that $L$ is a constant.
Now, if we had such an expression, we might also desire an inverse function $r(P)$. Indeed, this is what I desire. So the goal is to find an expression for $r(P)$.
My general approach so far has been to try and formulate $P(r)$ in such a way that $r(P)$ can be found algebraically. I have done this about 4 different ways now, and I will share with you just one as an example.

The area of the square which is covered by the circle $A$ can be represented as the area of four triangles plus the area of four arc sector areas.
The triangle areas can be represented as $ab = r\sin(\theta)r\cos(\theta) = \frac{1}{2}r^2\sin(2\theta)$, where $\theta = \cos^{-1}\left(\frac{L}{2r}\right)$.
The angle that forms the arc sector $\alpha$ can be written in terms of $\theta$ by observing its relationship with the right angle at the center. Therefore, $\alpha = \frac{\pi}{2} - 2\theta$.
From this we get the area of the arc sector as $\frac{1}{2}r^2\alpha = \frac{1}{2}r^2(\frac{\pi}{2} - 2\theta)$.
So the area of the square that is covered by the circle is given by $A(r) = 2r^2\sin(2\theta) + 2r^2(\frac{\pi}{2} - 2\theta) = 2r^2\sin(2\cos^{-1}\left(\frac{L}{2r}\right)) + 2r^2(\frac{\pi}{2} - 2\cos^{-1}\left(\frac{L}{2r}\right))$.
This makes the percentage of the square that is covered $P(r) = A(r)L^{-2} = L^{-2}(2r^2\sin(2\cos^{-1}\left(\frac{L}{2r}\right)) + 2r^2(\frac{\pi}{2} - 2\cos^{-1}\left(\frac{L}{2r}\right)))$.
Now the problem becomes apparent; If we want to solve for $r(P)$ we need to liberate $r$ from both $\sin(\frac{L}{2r})$ and $\sin(2\cos^{-1}(\frac{L}{2r}))$ while navigating around the $r^2$ terms.
This is the task I humbly put before you. My assumption is that there exists some formulation of the area which I have not yet considered which will be easily manipulated. Perhaps someone better versed in trig than myself can manipulate the existing formulation of the area into something better. Or maybe there is a way to formulate $r(P)$ directly.
 A: Based on the question of the percent of the square covered by the circle area was a bit hard to understand, however I will take that to mean the ratio of the squares area inside the circle divided by the squares area. This is easier to see if we divide the figures into eight pieces:
The only difference is that we are dividing each of the aforementioned areas by 8 which would simplify down to $\frac 88=1$. This leaves us with the area of the full triangle being $A_{triangle}$= $\frac 18 L^2$= $\frac 18 (\frac 85)^2$=$\frac {8}{25}$
First of all, the area in blue can be found by the right isosceles  triangle formula:$A_1=\frac 12 b^2=\frac 12(\frac{1}{\sqrt 2}) ^2$=$\frac 14$ while the area in green can be solved with an integral:
$A_2= \int_{\frac{1}{\sqrt 2}}^{\frac 45} \sqrt{1-x^2} dx$= $\frac 12 (x\sqrt{1-x^2}+sin^{-1}(x))$$|_{\frac {1}{\sqrt 2}}^{\frac 45}$=$\frac 12 sin^{-1}(\frac 45)-\frac {1}{100}-\frac π8$.
This means that the ratio is $P=\frac{A_1+A_2} {A_{triangle}} $100%= $\frac{\frac 14 + \frac 12 sin^{-1}(\frac 45)-\frac {1}{100}-\frac π8}{\frac {8}{25}}$100%=$[\frac {25}{16}tan^{-1}(\frac 43)-\frac {25π}{64}+\frac{3}{4}]$100%=97.171414789...%≈97.17% covered.
You could use the same technique here, but just have a different $A_1+A_2$ value. Let us have $A_{triangle}=A_▵$ These values should not be too hard to calculate exactly for a formula.
Here is a desmos demo with the general solution with the restriction of th $\frac{r}{\sqrt2}<inradius$ of the square=I<r:
https://www.desmos.com/calculator/9a9xnuwwxg
Here is the final answer for the restrictions above and the square inradius=R and the radius if the circle being r:
Final Answer
Set the slider in desmos for I=t and plot P. Ignore the errors and adjust the I value to get the function r=f(I) at least according to my formula.
Tell me if this helps or if I did something wrong!
