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At https://www.cut-the-knot.org/proofs/tessellation.shtml#solution it is said, "A sphere, like a plane, can be partitioned into a union of circles plus two points. Indeed, the two points can be chosen arbitrarily. They need not be the poles, say." Can you tell me how to do this?

I tried reducing it to the problem in the plane by stereographically projecting the two points onto the plane, constructing a coaxal family of nonintersecting circles that omit these two points, and then pulling the circles onto the sphere with the inverse projection, which takes circles to circles. But this doesn't cover the "north pole" on the sphere, so three points are omitted.

Can this method be patched up somehow to work, or is a different approach needed? It the latter, can you explain it, please?

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    $\begingroup$ Why don't you first use a rotation and send one point to the north pole....then do the stereographic projection...now your image is plane minus a point...cover it with disjoint circle...now pull back everything and inverse rotate the whole sphere. $\endgroup$ – Anubhav Mukherjee Oct 10 '17 at 19:00
  • $\begingroup$ I didn't try too much, but cannot you rotate your sphere so that one of your points end up on the north/do stereographic projection with one of your point as the pole ? $\endgroup$ – Junkyards Oct 10 '17 at 19:00
  • $\begingroup$ @Junkyards Thanks, but I don't see what you mean. If I rotate one point to the north pole, how do I pick the two points in the plane? $\endgroup$ – saulspatz Oct 10 '17 at 19:07
  • $\begingroup$ @AnubhavMukherjee Because then I will end up with the two omitted points being poles, if I understand you right. The two omitted points are supposed to be arbitrary. $\endgroup$ – saulspatz Oct 10 '17 at 19:10
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Given two non-antipodal points on the sphere, the tangent planes to the sphere at those points intersect in a line $L$ that doesn't intersect the sphere. Consider the other planes through $L$ that intersect the sphere. Their intersections with the sphere form your circles.

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  • $\begingroup$ Short and sweet. Thanks. $\endgroup$ – saulspatz Oct 10 '17 at 19:13
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See Apollonian circles at this link. Given two distinct points in the plane, the circles that pass through these two points form an elliptic pencil of circles. The family of curves orthogonal to this family is another pencil of circles. The solution is equivalent to the one given by @Robert Israel:. enter image description here

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  • $\begingroup$ But this is in the plane, not the sphere. I was familiar with this result, and was trying to use it to solve the problem on the sphere, but my approach didn't work. Perhaps I'm being dense, but I don't see how this is equivalent to Robert Israel's answer. Could you expand a bit, please? $\endgroup$ – saulspatz Oct 10 '17 at 23:07
  • $\begingroup$ @saulspatz: Well, how do you get the other pencil on the sphere: you consider all the planes through those two points and you take the circles cut by them on the sphere. They will be orthogonal to the other family. $\endgroup$ – orangeskid Oct 10 '17 at 23:11
  • $\begingroup$ Oh, I see what you're saying now, thanks. $\endgroup$ – saulspatz Oct 11 '17 at 3:57

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