I know about the inversion of a point inside a circle. But I was reading Peter Sarnak's paper on the Apollonian gasket, and got to the part where he was trying to prove descartes circle theorem. He described this circle inversion. Could someone please derive it?
1 Answer
$\begingroup$Sketch: Without loss of generality, assume the circle $E$ is centered at the origin. Inversion in $E$ is given in Cartesian coordinates by $$ (x, y) \mapsto (u, v) := \frac{k^{2}}{x^{2} + y^{2}}(x, y). \tag{1} $$ Particularly, $$ x^{2} + y^{2} = \frac{k^{4}}{u^{2} + v^{2}}. \tag{2} $$ The circle $C$ has equation $$ (x - d)^{2} + y^{2} = r^{2}. \tag{3} $$ Expand (3), substitute (2), and rearrange in the obvious way to get the equation of a circle. (Since you're reading Sarnak, this should be routine.)
If you're primarily interested in calculating the radius, put the center of $E$ at the origin and the center of $C$ at $(d, 0)$, so $C$ cuts the horizontal axis at $x_{1} = d + r$ and $x_{2} = d - r$. The image of $C$ under inversion crosses the horizontal axis at $$ x_{1}' = \frac{k}{x_{1}},\qquad x_{2}' = \frac{k}{x_{2}}; $$ the radius of the image is therefore $\frac{1}{2}|x_{2}' - x_{1}'|$, which is easily checked to be $$ \frac{k^{2}r}{d^{2} - r^{2}}. $$
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