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Trying to wrap my head around inversions. I understand it takes things from inside to outside, such that

$\text{distance from some point inside circle} + \text{ distance to new point}=r^2$

Where $r$ is radius of circle of inversion.

How would this look with a figure not just one line? For example a triangle inscribed in a circle? I know angles are supposed to be preserved.

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Let $O$ be the centre of the circle, $P$ our point "inside" and $Q$ the point it s mapped to. Then $(OP)(OQ)=r^2$. (Your expression in terms of sums is not quite right.) – André Nicolas Nov 16 '12 at 5:37
You might find this video enlightening: – Blue Nov 16 '12 at 6:18 – Rahul Nov 16 '12 at 6:23

Let $O$ be the centre of the circle, $P$ our point "inside" and $Q$ the point it s mapped to. Then $(OP)(OQ)=r^2$. (Your expression in terms of sums is not quite right.)

As to your question about the triangle, in general lines are taken to circles, except that lines through the origin are taken to themselves. So if the extensions of the lines that make up your triangle don't go through the origin, then the sides of the triangle will be sent to arcs of circles. These arcs will make up a vaguely triangular shape, but with circular arcs rather than straight lines. However, these arcs meet at angles that match the original angles of the triangle.

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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.)

Inversion of a vertical line!

  • 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.)

Inversion of a Bernoulli lemniscate

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