Let $\Pi$ be the plane whose intersection with the unit sphere $S$ is the spherical image of the circle $C$ in the plane with center $a$ and radius $r$, i.e., $\Pi \cap S = \sigma^{-1}(C)$. We can write the equation of $\Pi$ implicitly in the form ($\alpha_i \in \mathbb{R}$) (Ahlfors, p.19)
$ \alpha_1 x_1+\alpha_2 x_2 + \alpha_3 x_3 = \alpha_0, \quad \alpha_1 + \alpha_2 + \alpha_3 = 1 \quad 0 \leq \alpha_0 < 1, $
where $(x_1, x_2, x_3)$ are coordinate functions of the ambient space that the sphere $S$ lives in.
The circle $C$ corresponding to the intersection of this plane $\Pi$ and the unit sphere $S$ is given by (Ahlfors, p.19)
$ x^2 + y^2 - 2\frac{\alpha_1}{\alpha_0-\alpha_3}x+2\frac{\alpha_2}{\alpha_0-\alpha_3}y + \frac{\alpha_0}{\alpha_0-\alpha_3} - \frac{\alpha_3}{\alpha_0-\alpha_3} = 0 $
where $x$, $y$ are the real and imaginary parts of $z$, respectively. On the other hand, our circle has the given equation $\lvert z - a\rvert= r$, which written in coordinates $z = x + iy$ and $a = (a_1, a_2)$ has the form
$ (x-a_1)^2 + (y-a_2)^2 = r^2 \Rightarrow x^2 + y^2 - 2 a_1 x - 2 a_2 y + a_1^2 + a_2^2 - r^2 = 0 $
Equating the like powers of these two last equations, we can solve for $\alpha_0$, in particular, which turns out to be
$ \alpha_0 = \pm \frac{1 + \lvert a \rvert^2 - r^2}{\sqrt{\lvert a \rvert^4 - 2\lvert a \rvert^2(1-r^2)+(1+r^2)^2}} $
Now, our sphere $S$ is the unit sphere (the set of points that is distance 1 from the origin of $\mathbb{R}^3$) and the number $\alpha_0$ we just computed is the length of the line from the origin of $\mathbb{R}^3$ to the center of the circle $\sigma^{-1}(C)$ as provided in the Wikipedia article:
http://en.wikipedia.org/wiki/Plane%E2%80%93sphere_intersection
Thus from Pythagorean theorem, the desired radius $\rho$ of the circle $\sigma^{-1}(C)$ is given by
$ \boxed{ \rho = \sqrt{1-\alpha_0^2} = \Large{ \frac{2r}{\sqrt{\lvert a \rvert^4 - 2\lvert a \rvert^2(1-r^2)+(1+r^2)^2}}} } $