There's so many questions... Let me start with a useful hint.
We can rotate the curve $45$ degrees counter clockwise to get something we can better work with.
Rotation can be done with the use of Rotation matrix, which in this case gives us simply:
$$x=\frac{1}{\sqrt{2}} (y_2+x_2) \\ y=\frac{1}{\sqrt{2}} (y_2-x_2)$$
Now we just substitute the above expressions into the equation (and forget about subscripts right after):
$$\left| \frac{1}{\sqrt{2}} (y+x) \right|^{ \frac{1}{\sqrt{2}} (y+x)}=\left| \frac{1}{\sqrt{2}} (y-x) \right|^{ \frac{1}{\sqrt{2}} (y-x)}$$
We get the following plot (with circles added later of course):
The circumscribed circle really has radius $\sqrt{2}$, as the OP guessed. It can be seen by setting $y=0$ in the above equation.
The inscribed circle is more complicated. If we set $x=0$ we get:
$$\left| \frac{y}{\sqrt{2}}\right|^{ \frac{y}{\sqrt{2}}}=\left| \frac{y}{\sqrt{2}}\right|^{ \frac{y}{\sqrt{2}}}$$
But this is true for any $y$. Remember the line $y=x$ which satisfies the original equation? That's why.
By zooming in in Desmos we see that the radius is about $$a \approx 0.52026,$$
see below for the exact value.
Now, this kind of equation is very complicated to work with. However, we can use logarithms to make it more manageable:
$$(y+x) \log \left| \frac{1}{\sqrt{2}} (y+x) \right|=(y-x) \log \left| \frac{1}{\sqrt{2}} (y-x) \right|$$
This equation describes the same curve, but now we can work with it using analytical methods (such as Taylor series, in the appropriate ranges of $x$ and $y$).
For example, setting:
$$y>0 \\ y > |x|$$
We can get rid of the absolute values and write:
$$(y+x) \left( \log y+ \log \left(1+\frac{x}{y} \right)-\frac{\log 2}{2} \right)=(y-x) \left( \log y+ \log \left(1-\frac{x}{y} \right)-\frac{\log 2}{2} \right)$$
Simplifying:
$$2x \log y +y \left(1+\frac{x}{y} \right) \log \left(1+\frac{x}{y} \right)- y \left(1-\frac{x}{y} \right) \log \left(1-\frac{x}{y} \right) -x \log 2 =0$$
Now, to find the non-trivial root at $x=0$ (the radius of the inscribed circle) we can add the condition $y \gg |x|$ and expand the logarithmic functions, keeping only the terms up to $x^2$:
$$2x \log y +y \left(1+\frac{x}{y} \right) \left(\frac{x}{y}-\frac{x^2}{2y^2} \right)- y \left(1-\frac{x}{y} \right) \left(-\frac{x}{y}-\frac{x^2}{2y^2} \right) -x \log 2 =0$$
Simplifying (and keeping only the terms up to $x^2$), we get:
$$x \left(2 \log y+2 - \log 2 \right)=0$$
As we've already seen, $x=0$ gives us trivial solution for any $y$. However, if $x \neq 0$, but is very close to it, we have also another solution:
$$2 \log y+2 - \log 2=0$$
$$y=\exp \left( \log \sqrt{2}-1 \right)=\frac{\sqrt{2}}{e}=0.52026009502 \dots$$
This our $a$, radius of the inscribed circle.
There's actually another associated circle, which looks like this:
It is completely defined by the curvature of the function around $x=0$.
Its parameters could be deduced exactly (probably) by using the definition of the curvature, but I have used an approximation. Let's keep not only quadratic, but also cubic terms in the logarithm expansion. Then we get:
$$x \left(2 \log y+2 - \log 2-\frac{x^2}{3y^2} \right)=0$$
Rearranging, we can see that (for $x \neq 0$):
$$y=\frac{\sqrt{2}}{e} \exp \left(1+ \frac{x^2}{3y^2}\right) \approx \frac{\sqrt{2}}{e} \left(1+ \frac{x^2}{3y^2}\right)$$
From the general equation of the circle for small $x$ we have:
$$y=y_0-\sqrt{R^2-x^2} \approx y_0-R \left(1- \frac{x^2}{2R^2} \right)=y_0-R+\frac{x^2}{2R}$$
In our case, $R$ is a function of $y$, but let's make another approximation and assume $y= a=\frac{\sqrt{2}}{e}$, then we have:
$$R \approx \frac{3}{e \sqrt{2}} \\ y_0 \approx \frac{5}{e \sqrt{2}}$$
The diagram above uses these values and we can see they fit very well.
I'll add more to this once I think of something else.
Best Answer
As I understand, you have a curve of the form $$\tag{1} y_1 = \frac{100}{1+e^{-x}} $$ and a line $$ y_2 = 75 $$ Now, you want to calculate their intersection point. This can be done easily by hand. First, equate the two to get $$ y_1 = y_2 \qquad \Rightarrow \qquad \frac{100}{1+e^{-x}} = 75 $$ We can solve the $x$-coordinate quite easily. Multiply both sides by $\frac{1+e^{-x}}{75}$ to get $$ \frac{100}{1+e^{-x}}\left( \frac{1+e^{-x}}{75} \right) = 75 \cdot \left( \frac{1+e^{-x}}{75} \right) $$ As you probably see, a lot of the terms cancel out and we get $$ \frac{100}{75} = 1 + e^{-x} $$ We are close to the solution. Subtract $1$ from both sides to get $$ \frac{100}{75} -1 = e^{-x} $$ or $\frac{1}{3}= e^{-x}$. This equation can be solved by taking the natural logarithm on both sides. The result is $$ \ln{\left( \frac{1}{3} \right)} = -x $$ Remembering the properties of logarithms, we get $\ln{\frac{1}{3}} = - \ln{3}$ and therefore $$ x = \ln{3} \approx 1.0986 $$ Therefore, the coordinates are
You can test if this is the correct answer by inserting $x=\ln{3}$ into Equation (1)