Geometry - Solve a system of equations (maybe using WA or another CAS) I found this problem on the web, and it seemed like an interesting exercise to me.

So I am trying to solve this as a system of equations.
I tried using for example Wolfram Alpha.
That means e.g. I am trying to find the radius $R$ of the circle, and the point $x$, where the two curves touch each other.
I was hoping at least for a numeric solution. But I don't get anything  useful.
Could anyone help to get this working in WA?
I am not sure why WA doesn't understand this syntax.
Solve[Exp(-x^2) * (-2*x) == (R-x) / sqrt(x * (2*R-x)) && Exp(-x^2) == R + sqrt(x * (2*R-x)) && x > R && x < 2*R, {x, R}, Reals] 

How did I get these equations? Well, I think they should give me the solution to this problem above. Why? Because they express that
$f(x) = g(x)$  and also $f'(x) = g'(x)$
where the radius is denoted by $R$.
Here I have denoted
$$f(x) = e^{-x^2}$$
$$g(x) = R + \sqrt{x(2R-x)}$$
($g$ is the top semi-circle basically)
and I am looking for a solution $x$, where $x \in (R, 2R)$
I hope I didn't mess any calculations... but I think I didn't.
The two derivatives are easy to calculate.
So I am not sure why WA doesn't understand this.
EDIT:
I am not sure if any of the solutions below is actually an exact/explicit solution. I think I didn't get any formula for the blue area or for the touch point.
This is not a Mathematica stack exchange question actually. Mathematica was used just as a means to solve the problem. I am actually looking for a formula for the blue area (or for the X value of the touch point). I constructed a system of equations but solving by hand wasn't possible. Is there such formula, or do we have to use numeric approximations here? I think both answers here provide just a numeric approximation.
 A: HINT.- $(1)\space f(x)=e^{-x^2}$
$(2)$ The circle has radius $a$ and center $(a,a)$.
$(3)$ Tangent to this circle coincides with the tangent to $f(x)$ at some point $(x_0,y_0)$ so its equation is $\dfrac{y-y_0}{x-x_0}=-2x_0f(x_0)$.
$(4)$ Line perpendicular to this tangent allows us to finish.
Note.-You can have $a\approx 0.38$ (see the center pointed in attached figure)

A: Okay, this problem I initially thought was not that bad, but then I couldnt reason through all the logic and i got mad so i made it my personal vandetta to solve this using any method possible, so apologies if this is inefficient. You can probably salvage a faster method through this mess lol.
First, notice the circle is touching both coordinate axes, meaning its center will lie directly on the line $y=x$. WLOG let the equation of the circle be $$(x-r)^2+(y-r)^2=r^2$$
We can graphically see that we need the top part of the circle

Set this equal to our bounding equation so we have $$e^{-x^2}=\sqrt{r^2-(x-r)^2}+r$$
Now let's find another equation. If we draw the tangent line at the point where the circle meets the bounding equation we see that

By the distance formula, the distance from the point of tangency is $\sqrt{x^2+e^{-2x^2}}$. By adding up the radii, we see that this is equivalent to $r+r\sqrt2$. Setting these two equal to one another gives us a system of equations. (I just assumed the point the tangent point will lie on $y=x$, which is a good enough approximation. i will find an exact solution later)
\begin{align*}
e^{-x^2}&=\sqrt{r^2-(x-r)^2}+r\\
r+r\sqrt2&=\sqrt{x^2+e^{-2x^2}}
\end{align*}
Rewrite the second equation into $$r=\dfrac{\sqrt{x^2+e^{-2x^2}}}{1+\sqrt2}$$
and substitute it back into the first equation, we get (I did a FullSimplify on the substituted result first)
NSolve[((-1 + Sqrt[2]) (Sqrt[E^(-2 x^2) + x^2] + Sqrt[1 + Sqrt[2]] Sqrt[-(x (x + Sqrt[2] x - 2 Sqrt[E^(-2 x^2) + x^2]))]))==E^(-x^2), x]

$$x\approx0.65292...$$
Hence the radius is
N[Sqrt[0.65292^2 + E^(-2 0.65292^2)]/(1 + Sqrt[2])]

$$r\approx 0.38247...$$
And the area of the circle would be
$$\pi r^2\approx0.4595648...$$

In hindsight to approximate the point of tangency I literally could have done
Reduce[x==E^(-x^2)]

to get

which saves me from wasting all that computing time on the much more complicated reduce but whatever
