Suppose \(\ell\) is a line and \(C\) is a circle in the plane. Under what conditions is there a Möbius map which simultaneously sends \(\ell\cup\{\infty\}\) and \(C\) to concentric circles? Justify your conclusions.
Solution
1. A Möbius map sends generalized circles (circles or lines) to generalized circles.
Möbius transformations are conformal and preserve the class of generalized circles.
2. Two generalized circles can be sent to concentric circles iff they are disjoint (do not intersect).
2.1. Concentric circles are disjoint.
two distinct concentric circles have no common point.
2.2. Conversely, two disjoint generalized circles can be mapped to concentric circles.
a Möbius map sending a point of one circle to \(\infty\) turns it into a line; then a further map can be arranged so the two images are concentric circles (this is the standard fact that disjoint circles are “separated” and can be normalized to concentric circles).
3. Hence the condition is that \(\ell \cup \{\infty\}\) and \(C\) are disjoint.
3.1. \(\ell \cup \{\infty\}\) is a generalized circle (a line together with the point at infinity).
a line is a circle through \(\infty\).
3.2. The condition is that \(\ell\) does not meet \(C\) (in the finite plane).
\(\ell \cup \{\infty\}\) meets \(C\) iff \(\ell\) meets \(C\) (since \(\infty \notin C\) for a finite circle).
4. Therefore a Möbius map sending \(\ell \cup \{\infty\}\) and \(C\) to concentric circles exists iff the line \(\ell\) and the circle \(C\) are disjoint.
2 and 3.
5. Q.E.D.
4.