Circles generated by three-fold iterations $f(x)=\frac{1}{1-x}$

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 $$r(\phi)\approx1.93649$$

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. $$r(2)=r(-1)=r(\frac{1}{2})\approx1.76777$$)