We know that Mobius transformations, $z\to\frac{az+b}{cz+d}$, permutes circles and lines in the Euclidean plane, $(\mathbb{R}^2, dx^2 + dy^2 )$. 


It may even be possible to write an explicit formula for the general Mobius action on any given circle:


$$ |z-z\_0|=r \mapsto \left|\frac{az+b}{cz+d} -z\_0 \right|=r$$


Such a space can be generated by translations $z \mapsto z + z\_1$, rotations $z \mapsto \omega z$, dilations $z \mapsto rz$ and inversions $\displaystyle z = \frac{1}{\overline{z}}$. The action on all circles is clear except for the last case:


$$ |z-z\_0|=r \mapsto \left| z - \frac{\tfrac{1}{2}z\_0}{|z\_0|^2 - r^2} \right|= \frac{\tfrac{1}{2}r}{|z\_0|^2 - r^2} $$


Is there a more succinct way to write this transformation as a Lie group action? 


Here, the [Mobius group](https://en.wikipedia.org/wiki/M%C3%B6bius_transformation#Lorentz_transformations) $G = PSL(2, \mathbb{C}) \simeq SO(1,3)^+$ should act on on the space of circles (and lines) in the plane $\mathbb{R}^2$.




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This question also leads me to wonder what the space of circles in the Euclidean plane should be. Naively, the circles are a copy of $\mathbb{C} \times \mathbb{R}^+$ which looks like it could possibly be [Hyperbolic space](https://en.wikipedia.org/wiki/Hyperbolic_space) $\mathbb{H}^3$, in which case there would even be a natural metric.


However, the lines all have infinite radius. In which case, we still have three parameters, now a copy of $\mathbb{C} \times S^1$ identifying each line with the closest point to the origin and its direction. The $S^1$ behaving something like the [circle at infinity](https://www.youtube.com/watch?v=k-uWXIB7W7Y) (YouTube).


So now my space of circles is $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(\mathbb{C}\times S^1 \big)$. How does that happen?




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In fact this space should be $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(S^1 \times \mathbb{R}^+ \big)$.



Geometry of the space of circles in the Euclidean plane
Things will simplify if you just consider the circles on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$, for your space is simply the space of circles on the sphere (with the lines in $\mathbb{C}$ just becoming the circles through $\infty$). Thus, your extended space of circles becomes $M^3 = \mathrm{PSL}(2,\mathbb{C})/H$ where $H\simeq\mathrm{PSL}(2,\mathbb{R})$ is the subgroup of linear fractional transformations that preserve a given circle, say $\mathbb{R}\cup\{\infty\}$. (*See correction below. $\mathrm{PSL}(2,\mathbb{R})$ is actually the subgroup that preserves the oriented circle $\mathbb{R}\cup\{\infty\}$, when I actually need to divide by the $2$-component subgroup that preserves this circle but not necessarily its orientation.*)


Because $\mathrm{PSL}(2,\mathbb{R})\simeq\mathrm{SO}(2,1)^+$, it follows easily that this homogeneous space $M^3$ is naturally a Lorentzian manifold, but not naturally a Riemannian manifold.


If you want to think of $M$ as $\mathrm{SO}(3,1)^+/\mathrm{SO}(2,1)^+$, then you can see it as the space of oriented, Lorentzian $3$-planes in $\mathbb{R}^{3,1}$, i.e., it is identifiable with the quadratic hypersurface $\mathbb{S}^{2,1}\subset \mathbb{R}^{3,1}$ consisting of those vectors $v$ that satisfy $v\cdot v = 1$. The invariant Lorentzian metric on it is the one that it inherits as a submanifold of $\mathbb{R}^{3,1}$.


**Correction:** As Dylan points out, I should have divided by the subgroup $H\subset \mathrm{PSL}(2,\mathbb{C})$ generated by $\mathrm{PSL}(2,\mathbb{R})$ and the linear fractional transformation $f(z) = -z$, so that $H$ is the subgroup that preserves $\mathbb{R}\cup\{\infty\}$, but not necessarily its orientation. This means that the correct answer is $M = \mathrm{SO}(3,1)^+/H$, and this is actually realized not as $\mathbb{S}^{2,1}$ as described above, but as its quotient by the involution $v\mapsto-v$, which is actually $\mathbb{RP}^{2,1}\subset \mathbb{RP}^3$. Of course, it's still a connected Lorentzian $3$-manifold. (Whether some would call it "de Sitter space", I don't know.)


