circles and linear fractional transformations I'm realizing how little (in some respects) I know about circles.  Here's something that emerged out of something I was fiddling with.  My question is whether this is


*

*"well known" in the way that $229\times983=225107$ is "well known" (don't publish it unless you're publishing a table); or

*well known in the sense that every book includes it (for suitable values of "every"); or

*well known in the sense that everybody knows it (for at least moderately reasonable values of "everybody").


I'm looking at the circle $|z|=1$ in $\mathbb{C}$.  Let $$f(z)=\dfrac{-3z+1}{z-3}.\tag{This is $f$.}$$  This of course fixes $\pm 1$ and leaves the circle invariant, and maps $\pm i$ to $\dfrac{-3\pm4i}{5}$.  If we draw a circle through those two images of $\pm i$ meeting the unit circle at a right angle, it is centered at $-5/3$ and has radius $4/3$.  That circle meets the real axis at $-1/3$.  So look at the line $\operatorname{Re}=-1/3$.  Look at the point on that line where $\operatorname{Im} = y$.  Draw the line through that point and the aformentioned center $-5/3$.  That line crosses the circle twice.  It would seem that those two points are $f(z)$ and $f(-\bar z)$, where $z$ and $-\bar z$ are the two points on the unit circle with imaginary part $y$.
This gives us a simple geometric picture of how $f$ behaves.  That allows us to use routine Euclidean geometry to show that $\theta\mapsto f(e^{i\theta})$ satisfies the differential equation
$$
\left|\dfrac{dg}{d\theta}\right| = \text{constant}\cdot\operatorname{Re}\left(g-\left(-\dfrac 5 3\right)\right)
$$
subject to the constraint that the values of $g$ are on the unit circle.  (The equation says the rate at which $g$ moves along the circle is proportional to a certain affine function of the real part.)

Your differential equation follows from the well-known properties of Fractional Linear Transformations as conformal maps. 
Let me first do the specific case in your question: let 
$$ w = \frac{-3 z + 1}{z - 3} $$
we can solve
$$ z = \frac{1+3w}{w+3} $$
and hence
$$ \mathrm{d}z = \left(\frac{3}{w+3} - \frac{1+3w}{(w+3)^2} \right)\mathrm{d}w = \frac{8}{(w+3)^2} \mathrm{d}w $$
Now recall that 
$$ \left|\frac{\mathrm{d}w}{\mathrm{d}z}\right|  = \left|\frac{(w+3)^2}{8}\right| $$
is the conformal factor of the map $f: z\mapsto w$. That is to say, a unit tangent vector based at the point $z$ on $\mathbb{R}^2$, under the conformal map $f$, becomes a tangent vector based at $w$ on $\mathbb{R}^2$ with length $\frac{\left|w+3\right|^2}{8}$. 
Now, the object $\frac{\mathrm{d}g}{\mathrm{d}\theta}$ can also be written as $\partial f/\partial\theta$ and represents the image of the vector $\partial_\theta$ under the tangent map $\mathrm{d}f$. Since we are starting with $\partial_\theta$ tangent to the unit circle, it has unit norm, and hence we know that 
$$ \left| \frac{\mathrm{d}g}{\mathrm{d}\theta}\right| = \frac{ |g+3|^2}{8} $$
is precisely the conformal factor! Now using our knowledge that $g$ lies on the unit circle, we conclude that 
$$ |g+3|^2 = (\Re g + 3)^2 + (\Im g)^2 = (\Re g)^2 + 6 \Re g + 9 + 1 - (\Re g)^2 = 6 g + 10 $$
and hence we must have
$$ \left|\frac{\mathrm{d}g}{\mathrm{d}\theta}\right| = \frac{3}{4}\left( g + \frac{5}{3}\right) $$
as claimed. 
Note that this generalizes to any fractional linear transformation: given
$$ z = \frac{a w + b}{cw + d} \implies \mathrm{d}z = \frac{ac w + ad - acw - cb}{(cw+d)^2} \mathrm{d}w= \frac{ \begin{vmatrix} a & b \\ c & d\end{vmatrix}}{(cw+d)^2}\mathrm{d}w $$
which implies that 
$$ \left| \frac{\mathrm{d}w}{\mathrm{d}z} \right| = \frac{|cw + d|^2}{|ad - bc|}$$
Hence if $s\mapsto z(s)$ is a unit-speed curve such that $w(s) = f\circ z(s)$ (where $f$ is the inverse transformation to $w\mapsto z$ given above) lies on a circle centered at the origin, we have that necessarily 
$$ \left| \frac{\mathrm{d}w}{\mathrm{d}s} \right| = \frac{ |c|^2|w|^2 + |d|^2 + 2 \Re(\bar{d}c w) }{|ad - bc|} $$
and in particular solves an ODE of the type "speed is an affine function of the current coordinates" (And if $\bar{d}c \in\mathbb{R}$ we can even say that the "speed is an affine function of the current $x$ coordinate"). 
