Let $R > 0$ be a real number, and let $C = C_{R}$ denote the circle of radius $R$ centered at the origin. Inversion in $C$ is expressed in polar coordinates $(r, \theta)$ by $ (r, \theta) \mapsto (R^{2}/r, \theta). $ The geometric effect of inversion in $C$ can be explored pleasantly by looking at the polar graph $r = f(\theta)$, which maps to the polar graph $r = R^{2}/f(\theta)$.
Rotating and then inverting is the same as inverting, then rotating, so it's enough to consider a polar graph in some "standard angular position".
In the following examples, $c$ denotes an arbitrary positive real number.
- The line $x = c$ has polar equation $r\cos\theta = c$, or $r = c\sec\theta$. The image under inversion is the polar graph $r = (R^{2}/c)\cos\theta$, or $ x^{2} + y^{2} = r^{2} = (R^{2}/c)r\cos\theta = (R^{2}/c)x, $ which is well-known (and easily shown) to be a circle of radius $R^{2}/(2c)$ passing through the origin, with center on the ray $\theta = 0$. (To invert an arbitrary line $ax + by = c$ with $(a, b) \neq (0, 0)$, one can imagine rotating the plane through an angle $\phi$ to make the line vertical, inverting, then "un-rotating" by angle $-\phi$. The following pictures therefore suffice (in principle) to invert an arbitrary plane polygon not passing through the origin.)

- If $r^{2} = 2c^{2}\cos(2\theta)$ is a lemniscate of Bernoulli, the "figure-8" curve with Cartesian equation $ (x^{2} + y^{2})^{2} = r^{4} = 2c^{2} r^{2}\cos(2\theta) = 2c^{2}(x^{2} - y^{2}), $ the image under inversion is the square hyperbola with polar equation $(R^{2}/r)^{2} = 2c^{2} \cos(2\theta)$, i.e., $ (R^{2}/\sqrt{2}c)^{2} = r^{2} \cos(2\theta) = x^{2} - y^{2}. $ The tangents to the lemniscate at the origin correspond to the asymptotes of the hyperbola; here, to the lines $y = \pm x$. (A fancy way to say this is: In the Riemann sphere, a hyperbola looks like a figure-8, with the crossing point at infinity.)
