Everyone who discusses Möbius transformations mentions that they map circles to circles, but it can be hard to find concrete equations describing exactly how a given circle is changed by a given Möbius transformation.

I recently had to derive a formula for the curvature of a circle after applying a certain Möbius transformation, and the resulting formula was surprisingly simple. Suppose we start with a circle centered at \(x \in \mathbb{R}^2\) with radius \(r > 0\). If we apply a Möbius transformation which fixes the unit circle and sends some point \(z\) inside the unit disk to the origin, then the circle's curvature becomes \[ \tilde r = \frac{1-\|z\|^2}{1 + 2\langle x, z \rangle + (\|x\|^2 - r^2)\|z\|^2}r. \]

In particular, I was interested in Möbius transformations fixing the unit circle. If we ignore rotations, which don't affect the curvature of circles anyway, such a Möbius transformations is determined entirely by the point \(z\) which is sent to zero: \[f_z(p) := \frac{p-z}{1-\bar zp}.\] (See e.g. the wikipedia article.) Note that the inverse of \(f_z\) is given by \(f_{-z}\), since \(f_z(f_{-z}(p)) = p\).

Suppose we start with a circle of radius \(r\) centered at a point \(x \in \mathbb{R}^2\). We can encode our circle as the zero level set of a quadratic form \[Q(p) := a \|p\|^2 - \langle b, p \rangle + c = 0,\] with coefficients \(a = 1\), \(b = 2x\), and \(c = \|x\|^2 - r^2\). We can recover the center from \(Q\) as \(x = \tfrac c {2a}\), and the radius as \(r = \tfrac 1 {2a} \sqrt{\|b\|^2 - 4 ac}\).

Now we can determine how the Möbius transformation \(f_{-z}\) affects our circle by finding the zero set of the transformed map \(Q(f_{z}(p))\). The calculation is easier to do if we express \(Q(p)\) using complex numbers as \(Q(p) := a \bar p p - \Re(\bar b p) + c\). If we substitute in the definition of \(f_z(p)\) and do some nasty algebra, we find that \[\begin{aligned}\|1-\bar zp\|^2Q(f_z(p)) &= \left((a + \Re(\bar b z) + c \|z\|^2\right) \|p\|^2\\ &\quad- \Re\left((2 a \bar z + \bar b + b \bar z^2 + 2 c z)p\right)\\ &\quad+ (a \|z\|^2 + \Re(\bar b z) + c) \end{aligned}.\]

That is, up to a scalar multiple, \(Q(f_z(p))\) is itself a quadratic form with coefficients \[\begin{aligned} \tilde a &:= a + \Re(\bar b z) + c \|z\|^2\\ \tilde b &:= 2az + b + \bar b z^2 + 2 c \bar z\\ \tilde c &:= a \|z\|^2 + \Re(\bar b z) + c. \end{aligned}\]

We can now extract the transformed radius from \(Q(f_z(p))\). Another gnarly calculation shows that \[ \|\tilde b\|^2 - 4 \tilde a \tilde c = (\|b\|^2 - 4 a c)\left(1-\|z\|^2\right)^2 = 4a^2 r^2 \left(1-\|a\|^2\right)^2. \] Therefore, the radius of the transformed circle is given by \[ \tilde r = \frac{1-\|z\|^2}{1 + \tfrac 1a\langle b, z \rangle + \tfrac ca \|z\|^2}r, \] as promised above. (The sign on the \(\langle b z\rangle\) term is flipped in the earlier expression above since there we applied \(f_z\) to the circle, rather than \(f_{-z}\).)

The \(1-\|z\|^2\) term suggests that there may be a slick proof of this formula using hyperbolic geometry: Möbius transformations which fix the unit circle are precisely the isometries of the Poincaré disk model of the hyperbolic plane, whose metric tensor is given by: \[ds^2 = \frac{4 dz^2}{\left(1-\|z\|^2\right)^2}.\] But I'm happy with this concrete calculation for now.