Why is inversion in a straight line the same as reflection? In the inversive plane points $P$ and $P'$ are defined to be inverses with respect to a circle $\omega$ of radius $k$ and center $O$ when they are distinct from $O$, on the same ray emanating from $O$, and $OP * OP' = k^2$. When $\omega$ is a straight line (or circle of infinite radius), the operation of inversion is the same as a reflection (across the line). However, since the equation $OP * OP' = k^2$ breaks down with $O$ becoming the point at infinity when $\omega$ is a straight line, I am having trouble seeing the precise argument that is made to justify the equivalence of inversion and reflection in this case. Can anyone supply this argument?

This is a tricky problem, where the real question is, "Why should we define it as ..." rather than "Prove it is...." We can give motivation for the definition.
The first justification is continuity.
Let $P$ be a point on the line, and let $v$ be a a unit vector perpendicular to the line. For any radius, $r$, let $O_r$ be the point $P+rv$. Then the inversion of the circle with center $O_r$ and radius $r$ is a function on the plane. As $r\to\infty$, that function has, as a point-wise limit, the reflection through the line. 
There isn't a single "point at infinity" in this case, but different points depending on the line's slope (specifically, different points at infinity, one per unit vector $v$.) That indicates that what we are looking at is something related to the projective plane, but I'm failing to recall a projective view of the inversion operation...
