TITLE: Circles generated by three-fold iterations $f(x)=\frac{1}{1-x}$
QUESTION [15 upvotes]: I came across a weird property of the function $f(x)=\dfrac{1}{(1-x)}$
Observe the following: 
$$f(x) = \frac{1}{(1-x)}, \quad\quad f^2(x) = f(f(x)) = \frac{(x-1)}{x}, \quad\quad
f^3(x) = f(f(f(x))) = x$$ 
ultimately implying that $f^2(x)=f^{-1}(x)$.

(Mini question: Do you know of any other functions $g(x)$ where $g \circ (g \circ g(x)) = g^3(x)=x$ aside from $f(x)$ and aside from the trivial case where $g(x)=x$? I was pretty shocked when I noticed this pattern with $f(x)$.)


Anyway, notice for every $x$, there is a set of triplets generated by repeatedly applying the function $f(x)$. 
Specifically $\langle x\rangle =\{x,f(x),f^{-1}(x)\}=\{x,\frac{1}{(1-x)},\frac{(x-1)}{x}\}$ 
For an illustrative example let $x=2$, so then $\langle 2\rangle=\{2, -1, \frac{1}{2}\}$. See now that this can be thought of as 3 points on the graph of the function $f(x)$, where
Point $A$: $x \mapsto f(x)$
Point $B$: $f(x) \mapsto f^{2}(x)=f^{-1}(x)$
Point $C$: $f^{-1}(x) \mapsto x$

Explicitly, still using $x=2$ as the example: 
Point $A$: $(x, f(x)) = (2,-1)$
Point $B$: $(f(x), f^{-1}(x)) = (-1,\frac{1}{2})$
Point $C$:     $(f^{-1}(x),x) = (\frac{1}{2},2)$

OK so now my question! 

Since 3 points uniquely define a circle, I'd like to know if we can derive a closed-form function $r(x)$ that calculates the radius of circle $R$, where circle $R$ is the circle uniquely defined by the 3 points $A$, $B$ and $C$ generated by $\langle x\rangle$. 

Continuing the example where $x=2$, circle $R$ has center at Point $R=(\frac{3}{4},\frac{1}{4})$ (i.e. the circumcenter of points $A$, $B$ and $C$). The radius of circle $R$ is then simply:
$$|\overline{AR}|=\sqrt{{\left(2-\frac{3}{4}\right)}^2+{\left(-1-\frac{1}{4}\right)}^2}= \frac{5\sqrt{2}}{4}.$$
So evaluating $r(x)$ at $x=2$ gives us $r(2)=\dfrac{5\sqrt{2}}{4}\approx1.76777$.

Another cool example to consider is $x=\phi$, where $\phi=\dfrac{1+\sqrt{5}}{2}\approx1.61803$ (the Golden Ratio). Some cool characteristics that make $\phi$ unique among all numbers are:
$$\phi-1=\frac{1}{\phi}\quad\text{and}\quad \phi+1=\phi^2$$
You can calculate this on your own, but applying $f(x)$ on $x=\phi$ repeatedly results in $\langle\phi\rangle=\{\phi,-\phi,\frac{1}{\phi^{2}}\}$.
With the help of Wolfram Alpha, I was able to calculate $r(\phi)\approx1.93649$
(Circumcenter: https://tinyurl.com/y59trfn5 | Radius: https://tinyurl.com/y6jxs9sn)

Calculating the circumcenter seems to be the biggest issue, but maybe there's a cleaner way with the help of linear algebra? I was reading that there's a way to calculate the formula of a circle using matrices and determinants, but that seemed too complex for this. Maybe circles and triangles aren't the way to approach this at all -- I'd be happy to take suggestions and hear your thoughts!
Just some last conceptual thoughts...
1)  $r(x)$ should always be positive (i.e. there is no $x$ where $r(x)$ is $0$ or negative), and therefore somewhere hit some positive minimum value for $r(x)$ (assuming/implying that $r(x)$ is smooth and differentiable on the interval $x \in (-\infty,1)\cup(1,+\infty)$).
2) $\lim\limits_{x \to 1^-}r(x)=+\infty$ and $\lim\limits_{x \to 1^+}r(x)=+\infty$
3) $\lim\limits_{x \to -\infty}r(x)=+\infty$ and $\lim\limits_{x \to +\infty}r(x)=+\infty$
4) $r(x)$ is NOT symmetric around $x=1$. Just as a quick check, $r(3)\approx2.12459$ and $r(-1)\approx1.76777$
5) $r(x)$ is actual VERY NOISY as a function since for any 1 value of $r(x)$, there are at least 3 unique variables that result in that value (i.e. all $x \in \langle x\rangle$)(e.g. $r(2)=r(-1)=r(\frac{1}{2})\approx1.76777$)

That last point makes me feel there's no true closed-form function for $r(x)$. Regardless, I'd be really curious to find out what the minimum radius is... (placing \$1 on $r(x)$ for $x \in \langle\frac{\pi^2}{4}\rangle$!)

REPLY [6 votes]: The equation of a circle through $A$, $B$, $C$ is given by:
$$\left|\begin{array}{cccc}
x^2+y^2 & x & y & 1 \\
A_x^2+A_y^2 & A_x & A_y & 1 \\
B_x^2+B_y^2 & B_x & B_y & 1 \\
C_x^2+C_y^2 & C_x & C_y & 1 
\end{array}\right| = 0 \tag{1}$$
For
$$A = \left(p,\frac{1}{1-p}\right)\qquad B = \left(\frac{1}{1-p},\frac{p-1}{p}\right) \qquad C = \left(\frac{p-1}{p},p\right)$$
this becomes (with the help of a computer algebra system)
$$\left(x^2+y^2\right)p(p-1) - x\left(1- 3 p + p^3\right) -y\left(1 - 
 2 p - p^2 + p^3\right)  = -1 - p + 4 p^2 - p^3 \tag{2}$$
Completing the square yields
$$\left(x -\frac{p^3-3p+1}{2p(p-1)} \right)^2+\left(y - 
\frac{p^3-p^2-2p+1}{2p(p-1)} \right)^2 = \frac{\left(p^2 + 1\right) 
\left(p^2-2p+2\right) \left(2p^2-2p+1\right)}{4 p^2(p-1)^2} \tag{$\star$}$$
Thus, the circumcenter $K$ and radius $r$ are

$$\begin{align}
K &= \left(\frac{p^3-3p+1}{2p(p-1)}, \frac{p^3-p^2-2p+1}{2p(p-1)}\right)\\[6pt]
r &= \frac{\sqrt{\left(p^2+1\right) \left(\,(p-1)^2+1\,\right) \left(\,p^2+(p-1)^2\,\right)\;}}{2 |p(p-1)|}
\end{align}$$

Interestingly, we can rewrite $r$ as
$$r = \frac12 \;\sqrt{\left(\frac{p}{1} + \frac{1}{p}\right) \left(\frac{p - 1}{1} + \frac{1}{p - 1}\right) \left(\frac{p}{p - 1} + \frac{p - 1}{p}\right)}$$
Animation courtesy of @Jyrki:


By way of minimization ... We have
$$\frac{d(r^2)}{dp} = \frac{(p-2) (p+1) (2p-1) (p^2-p+1)^2}{2 p^3(p-1)^3}$$
so that the non-extraneous critical points occur at $p=2,-1,\frac12$, which correspond exactly to OP's $\langle 2\rangle$ triangle, as can be seen in the animation.