As is well-known, lines and circles are converted into lines and circles by circle inversion, or by any Möbius transformation for that matter. What bothers me is what happens in the Pappus's classical construction of a chain of circles inscribed into a region between two tangent circles (so-called Archimedes's arbelos).

One can apply a strategic circle inversion that converts this region into a strip between two parallel lines (see the first image on Mathhelp's Pappus Chain and below). Naturally, the inscribed circles are inverted into a vertical stack of circles inscribed into the strip, and their centers lie on its midline. The trouble is that the centers of the original Pappus chain circles lie on an ellipse (this is easy to show using its focal property, see e.g. Wikipedia's Pappus Chain). Since the inversion is involutive it would seem that it inverted a line into an ellipse?!

I am probably missing something very simple but I am not sure what. Is it that circle centers are not inverted into circle centers? Where does this ellipse go then?

enter image description here

  • 1
    $\begingroup$ That’s it exactly. The image of a circle’s center under inversion is usually not the center of the circle’s image. $\endgroup$ – amd Jan 26 '18 at 0:00
  • $\begingroup$ @amd But it is hard to see where the ellipse could go other than the midline given the symmetry of the pictures. The inversion of the midline would have to be the circle through all the tangency points, and I am starting to think that perhaps the ellipse of centers does not pass through those despite how it looks. $\endgroup$ – Conifold Jan 26 '18 at 0:09
  • $\begingroup$ Inversions of ellipses yield some odd-looking curves. $\endgroup$ – amd Jan 26 '18 at 0:24

Your guess is correct. Circle centers aren’t mapped onto circle centers.

For simplicity, consider inversion in the unit circle. Let $c$ be the radial distance of the center of a circle of radius $r$ to be inverted. The ray through this point intersects the circle at radial distances of $c\pm r$. These two points are inverted to $(c\pm r)^{-1}$. Their midpoint—the center of the circle’s image—is at ${c\over c^2-r^2}$, but the image of the circle’s center is at $\frac1 c$.

For the Pappus chain, the ellipse on which the centers lie has the centers of the outer and inner circles that generate the chain as its foci. The inversion circle is orthogonal to the first circle of the chain (the circle that completes the arbelos). Taking, for example, an outer diameter of $7$ and inner diameter of $4$, the central ellipse has semi-axis lengths $\frac{11}4$ and $\sqrt7$. The inversion circle has a radius of $2\sqrt7$, so the inversion formula is $\mathbf r'=28{\mathbf r \over \|\mathbf r\|^2}$. With some simplification help from Mathematica, the inversion of the central ellipse is $$x = {1232 \over 233+ 9\cos t}, y={448 \sqrt{7} \tan \frac t2 \over 233+ 9\cos t},$$ the red curve in the following illustration.

Pappus chain with inverted central ellipse.

This curve asymptotically approaches the vertical centerline of the inverted circle stack, so as you get farther along the chain, the inverted circle centers do almost lie on a straight line, but none of them actually lie on it. With Mathematica’s help again, the parameter $t$ can be eliminated to get the implicit equation $$2\arctan{11y \over 4\sqrt7 x} = \arccos{1232-233x\over9x},$$ but I don’t find that particularly illuminating.

  • $\begingroup$ Computation in polar coordinates shows that the midline is inverted into the circle through tangency points, which does not pass through centers, and the ellipse of centers does not pass through tangency points. Its inversion hangs around a vertical line, slightly bent to the left near the $x$-axis and to the right at infinities. But the vertical line is strictly to the left of the midline, which surprised me. Deviations are regulated by the difference between arithmetic and harmonic means of the radii so when arbelos is drawn with radii not too far apart they are visually undetectable. $\endgroup$ – Conifold Jan 26 '18 at 19:25
  • $\begingroup$ @Conifold The center of the first Pappus circle lies just outside of the inversion circle, so that center will certainly end up to the left of the midline. $\endgroup$ – amd Jan 26 '18 at 19:39
  • $\begingroup$ This is a great post! One question though... do the circle centers of the "next order" Pappus chain also form an ellipse? (i.e. the chain formed by drawing circles tangent to the blue circles and the black circle on the left) If so, how can we show that? $\endgroup$ – Math Enthusiast Dec 24 '18 at 17:16
  • $\begingroup$ @MathEnthusiast That’s worthy of a separate question of its own! $\endgroup$ – amd Dec 24 '18 at 18:09
  • $\begingroup$ Sorry haha :) I think I just figured it out! Given that for any chain of circles enclosed by two (non-concentric) circles, the centers of the circles form an ellipse (by the same reasoning as above), and given that the black circle on the left already forms the inner bound, all we have to do is prove that there exists some outer circle which bounds the chain. This is trivial, because any Pappus chain of order n forms a linear chain of circles in the inversion, and is therefore bounded by two lines. Invert back, and the chain is bounded by two circles, therefore the centers form an ellipse! $\endgroup$ – Math Enthusiast Dec 28 '18 at 18:35

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