How do you work with the space of circles on the sphere considered as the projective line? I'm trying to prove some things about the action of the Möbius group on the "circlines" in the extended complex plane, ie. circles on $\mathbb{C}P^1$.
I find that while I have a good grip on Möbius transformations (as $PSL(2, \mathbb{C})$), I don't really know how I'm supposed to treat circles.
Points on the sphere are lines in $\mathbb{C}^2$, so I'm wondering if there's a way to represent circles as things in $\mathbb{C}^2$. More concretely, let $X$ be a circle in $\mathbb{C}P^1$, and $\pi : \mathbb{C}^2 \setminus 0 \to \mathbb{C}P^1$ the quotient map. What is $\pi^{-1}(X)$?
EDIT: Many thanks to John Hughes for helping me figure out some of the earlier parts of my question:

Abstractly speaking, is the space of circles on the sphere a manifold?
  Or maybe some sort of space with singularities? What dimension is it?
  How do I work with it?

It seems that circles on the sphere do in fact form a manifold which is double covered by $S^2 \times (-1,1)$ or $S^2 \times [-1,1]$ depending on whether degenerate circles (=points) are allowed. The space of circles is one of the above manifolds quotiented by the relation $(x,y) \sim (-x,-y)$. The quotient of the former becomes a manifold, and the quotient of the latter becomes a manifold with boundary.
 A: Assuming that you're either talking about great circles or circles of nonzero radius, then yes, the space of circles on a sphere is a manifold. 
To make sure I don't screw things up, I'm going to talk about the case of great circles on a real sphere, because that has all the features you care about. In this case, each great circle corresponds nicely to a normal line to the plane of the circle, so you've got a correspondence between the space of great circles and $RP^n$. If the circles are oriented, then you get a correspondence with $S^n$. 
For all circles (but I'll restrict to nonzero radius, and oriented circles), you have the following correspondence: your circle lies in an oriented plane with an oriented unit normal vector $v$. For any point $P$ of the circle, $ t = (P - C) \cdot v$ tells how far this plane is from the sphere's center, $C$. Alternatively, given $v$ and $t$, you can reconstruct the circle. So this gives a 1-1 correspondence between your set of circles and $S^2 \times (-1, 1)$. 
$$
\newcommand{\ii}{\mathbf i}
$$
Your central question: if $X$ is a circle in $S^2 = \mathbb CP^1$, and points of $S^2$ correspond to lines in $\mathbb C^2$, what does a circle in $S^2$ correspond to? Well...to a family of lines in $\mathbb C^2$. Let's look at the equator as an example. Under stereographic projection from the north pole, this corresponds to the unit circle in $\mathbb C$. Each point $z = \cos t + \ii \sin t$ of the circle corresponds to the line in $\mathbb C^2$ that passes through the point $(z, 1)$. So the preimage of this circle, in 
$\mathbb C^2 \setminus 0$, is $$\{ (  a\cos t + \ii a\sin t, a) \mid t \in \mathbb R, a \in \mathbb C \setminus 0 \}.$$
This is sort of a "double cone" in $\mathbb C^2$, with its apex at the origin, except that each of the ruling lines or "generators" of the cone is a complex line (i.e., a plane) rather than a 1-dimensional line. 
A: Theorem. A circline of $\mathbb{C}$ corresponds in $\mathbb{CP}^1$ to the zero locus of a nonzero hermitian quadratic form on $\mathbb{C}^2$ (defined up to scalar multiples) with $\det\le0$ (the degenerate circlines correspond to $\det=0$).
Indeed, for the quadratic equation $|w-p|^2=r^2$ (where $p$ is the center and $r$ the radius), we may replace $w$ with $w/z$ to work projectively (recall $[w:z]$ in $(\mathbb{C}^2)^\times/\mathbb{C}^\times=\mathbb{CP}^1$ corresponds to $w/z$ in $\widehat{\mathbb{C}}=\mathbb{C}\sqcup\{\infty\}$), and rewrite $|w-pz|^2=r^2|z|^2$ (using $|z|^2=\overline{z}z$) as $\overline{w}w-p\overline{w}z-\overline{p}w\overline{z}-r^2\overline{z}z=0$, or in matrix form
$$ \begin{bmatrix} w \\ z \end{bmatrix}^\dagger
\begin{bmatrix} 1 & p \\ \overline{p} & |p|^2-r^2 \end{bmatrix}
\begin{bmatrix} w \\ z \end{bmatrix}=0 $$
Thus, the circle $C(p,r)$ corresponds to the hermitian matrix $\left[\begin{smallmatrix}1&p\\ \overline{p}&|p|^2-r^2\end{smallmatrix}\right]$ (or scale by $\frac{1}{r}$ to get $\det=-1$). It may not be clear what to do for actual lines, but notice circlines (circles and lines) in $\mathbb{C}$ correspond in $\mathbb{CP}^1\simeq S^2$ via stereographic projection to just all circles (lines correspond to circles through the north pole, the point of projection). Recall the formulas for stereographic projection are
$$ [\alpha:\beta] \mapsto \left(\frac{\mathrm{Re}(\alpha\overline{\beta})}{|\alpha|^2+|\beta|^2},\frac{\mathrm{Im}(\alpha\overline{\beta})}{|\alpha|^2+|\beta|^2},\frac{|\alpha|^2-|\beta|^2}{|\alpha|^2+|\beta|^2}\right), \quad (x,y,z)\mapsto\frac{x+yi}{1-z}. $$
Consider a spherical circle with center $\vec{x}=(x,y,z)\in S^2$ and spherical radius $\theta$. (Note the stereographic projection of $u$ generically does not correspond to the center of the stereographcially projected circle.) The points $\vec{u}=(u,v,w)$ on the spherical circle $SC(\vec{x},\theta)$ are defined by the dot product $\vec{u}\cdot\vec{x}=\cos\theta$.
Exercise. Show the $\vec{u}\in SC(\vec{x},\theta)$ correspond to $[\alpha:\beta]\in\mathbb{CP}^1$ which is the zero locus of the hermitian quadratic form defined by the matrix $\left[\begin{smallmatrix} z-\cos\theta & x+yi \\ x-yi & -z-\cos\theta \end{smallmatrix}\right]$, where $\vec{x}=(x,y,z)$.
Denote $Q_0=\left[\begin{smallmatrix} z-\cos\theta & x+yi \\ x-yi & -z-\cos\theta \end{smallmatrix}\right]$ and $Q_\theta=Q_0-(\cos\theta)I_2$. Defining $n(\vec{x})=\left[\begin{smallmatrix} x+yi \\ 1-z\end{smallmatrix}\right]$, the eigenvalues of $Q_0$ are $\pm1$ with respective eigenvectors $n(\pm\vec{x})$. When $\sin\theta=0$ (i.e. $\cos\theta=\pm1$), $Q_\theta$ is (up to a scaling factor of $2$) a projection onto $n(-\vec{x})$, and otherwise we can normalize $\hat{Q}=(\csc\theta)Q_\theta$ in which case $\det\hat{Q}=-1$ and the eigenvalues are $\tan(\theta/2)$ and $-\cot(\theta/2)$ by half-angle formulas.
Importantly, every nonzero hermitian matrix with $\det\le0$ corresponds (up to scaling) to exactly one (possibly degenerate) circle on the Riemann sphere. The effect of Mobius transformations corresponds to similarity transformations of the hermitian matrices (exercise - why?).
