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.

Image of a Möbius transformation from Möbius transformations revealed by Arnold & Rogness.

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.