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I've been looking for solutions to finding the set of circles tangent to two other circles. one circle can be inverted to a line, but two circles can be mapped to a line and a circle

or equivalently one circle tangent to three circles

has the Apollonius problems been approached from intersection theory. in this case CCC

Are they solvable from that perspective? this looks like a very classical algebraic geometry problem about Mobius geometry.

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  • $\begingroup$ Usual approach is at first to make one of radii equal to 0 (by subtracting the minimal radius from all radii), after that make inversion in this point (degenerated circle). The problem reduces to a common tangent to two circles, that is easy. $\endgroup$ – Fedor Petrov Jul 16 '16 at 22:15
  • $\begingroup$ how to get the general case from radius equals 0. there must be a few solutions buried in the literature. or nice modern discussions of quadriceps somewhere $\endgroup$ – john mangual Jul 16 '16 at 22:31
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    $\begingroup$ A version also posted @MSE, "approaches to Apollonius circle problems" (by user cactus314). $\endgroup$ – Joseph O'Rourke Jul 17 '16 at 2:26
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Two circles with different radii can also be mapped to two circles with the same radius by an inversion (or a möbius transformation in general). Then the family of tangent circles is easier to describe. Similarly one could send three circles of different radii to three circles of the same radii by an inversion (möbius also in general), as long as the circles are not too far apart from each other (this can be properly formulated).

In general the closest thing to what you are describing seems to me is the solution to the Appolonius problem via the methods of Lie sphere geometry (simply google it and you will find it). The idea is that the set of oriented circles (points being treated as circles of radius zero) lying in the two sphere form a 3 dimensional quadric in $\mathbb{P}(\mathbb{R}^{5,2})$ with a dot product of signature $+++--$. Two oriented circles are tangent when the dot product between their corresponding points in the model is zero (i.e. the vectors representing the points are orthogonal). Then three oriented circles if in general position form three points on the quadric, determining a projective two dimensional plane. The polar projective subspace to that plane with respect to the quadric (which is the orthogonal complement before projectivization) is a projective line. Depending on its signature, the line can intersect the Lie quadric in zero, one or two points, which are the solutions to the Appolonius' problem. Observe that the orientation of the circles is crucial. Möbius geometry is a sub geometry of Lie sphere geometry. I hope this helps.

Edit: Since you asked about the relation between the two nested circles (red and blue) and the number of black circles in the Steiner chain, here is a relation, if I have not made a mistake in my computations: Let $R$ and $r$ be the radii of the larger circle and the smaller circle inside it respectively. Let $d$ be the distance between the centers of these two circles. Then one gets an $n$ chain that wraps around $m$ times whenever $$ \frac{R(d+r-x)}{(d+r)x-R^2} =\left( \sec\left(\frac{m}{n} \pi\right)+\tan\left(\frac{m}{n} \pi\right)\right)^2,$$ where $x$ is the larger solution of the quadratic equation $$ (R^2-xd)(x-d)-r^2x=0. $$

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  • $\begingroup$ is this like the Veronese embedding ? $\endgroup$ – john mangual Jul 17 '16 at 3:36
  • $\begingroup$ Yeah, the philosophy is somewhat similar, when you think of Veronese embedding as a tool for studying incidence between points and conics... I guess... $\endgroup$ – Futurologist Jul 17 '16 at 4:10
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Perhaps you are looking for a Steiner chain?


         
          (Images (above & below) from Wikipedia.)
          Chain Animated


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  • $\begingroup$ totally if the Blue circle is the unit circle, which radii allows me to have $n$ circles? in principle this can be computed by taking a mobius transformation of the regular case but I still have difficulty computing the centers and radii $\endgroup$ – john mangual Jul 17 '16 at 3:35
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    $\begingroup$ I don't know if it helps but these chains can be transformed to Poncelet's porism: the points of tangencies between pairs of black circles lie on a common circle orthogonal to all black ones. Furthermore the centers of the black circles lie on an ellipse. So the poligonal line formed by connecting pairs of centers of adjacent black circles is inscribed in the ellipse and tangent to the orthogonal circle. $\endgroup$ – Futurologist Jul 17 '16 at 4:49
  • $\begingroup$ @johnmangual: There is a Mathematica notebook linked from here. For a nice animation by Steven Dutch of the circles "rolling," see this link. $\endgroup$ – Joseph O'Rourke Jul 17 '16 at 18:23

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