# Inversive distance between concentric circles

I found an explication about Inversive distance in Coxeter book. But, i don't really understand some things in its explication. I can't imagine the meaning of:

''This inquiry almost forces us to invent, for any two non-intersecting circles $$\alpha$$ and $$\beta$$, an inversive distance ($$\alpha$$, $$\beta$$) such that, if $$\gamma$$ belongs to the non-intersecting pencil $$\alpha\beta$$ and if $$\beta$$ lies between $$\alpha$$ and $$\gamma$$, then $$(\alpha, \beta) + (\beta, \gamma) = (\alpha, \gamma)$$".

Then it says:

"Inverting in a circle whose center is one of the limiting points, we obtain three concentric circles."

I do not understand the meaning of "circle whose center is one of the limiting points".

Here is an image of that page.

Here is what i get but only 2 concentric circles.

## 1 Answer

For two circles $$\alpha$$ and $$\beta$$ with radical axis $$l$$, the pencil $$\alpha\beta$$ is the set of circles $$\gamma$$ which share this radical axis, i.e. the radical axis of $$\gamma$$ and $$\alpha$$ is $$l$$ as well (and thus automatically also the radical axis of $$\gamma$$ and $$\beta$$). In the picture below, the pencil is drawn in solid lines (black), and the common axis is the vertical (grey) line.

If and only if the circles do not intersect, there exists a circle (dashed red in the picture), which is centered at the intersection of $$l$$ and the common axis of symmetry $$g$$ of the pencil, and orthogonal to all circles of the pencil. This circle intersects the axis of symmetry in the two limiting points of the pencil. These can be thought to be the two circles of zero radius belonging to the pencil.

Inverting in a circle (dotted blue in the picture) centered at one of those points will leave the axis of symmetry invariant. The dashed circle, however, becomes a line orthogonal to $$g$$, since it passes through the center of inversion. A circle belonging to the pencil will be transformed to another circle orthogonal to both lines, which is thus necessarily centered at their intersection. Therefore, the pencil is transformed into a pencil of concentric circles, all centered at the intersection of $$g$$ and the image of the dashed circle.

Since the radius of the circle of inversion is arbitrary, and choosing another radius corresponds to scaling the inverted picture, any well-defined quantity derived from the inverted picture has to be invariant under scaling. Therefore, it has to be defined as an expression of the ratios of the concentric circles - which is exactly what Coxeter proceeds to do. He defines $$(\alpha, \beta)$$ - the "distance" of the two circles - to be the logarithm of the ratios of the radii of the inverted circles.

• Sorry for stupid question, but how is it possible that your circle of inversion has the same center as 2 inversed circles ? I don't really understand how we get 3 concentric circles but not only 2. I added an image of what i get in my question. Thank you! @Josef E. Greilhuber – Daniil Aug 18 '19 at 11:30
• @Daniil I don't really see the problem. Draw any third circle of the coaxial pencil, invert it, and you will get a third concentric circle in your picture. The circle of inversion is not concentric with the circles of the pencil, it just really looks like that in my picture (I hope I didn't make a mistake drawing it). Is that what you meant? – Josef E. Greilhuber Aug 18 '19 at 12:01
• @Daniil Oh yes, I might have accidentally drawn the second smallest circle of the pencil with the wrong center (they are awfully close together). I will correct it as soon as I am at a reasonable computer. – Josef E. Greilhuber Aug 18 '19 at 12:04
• This is a definition, not a proof. That it's base $e$ doesn't really matter, you could use any other base (just as you could measure distances in feet instead of metres). Importantly, while $(\alpha,\beta)$ is called the "inversive distance" of circles $\alpha$ and $\beta$, it really is not the euclidean distance, since it is scale-invariant! It just behaves like a distance function of two objects "should" behave, in that it satisfies the triangle inequality, with equality if and only if the three circles are in a nice configuration (the pencil in this case). – Josef E. Greilhuber Aug 19 '19 at 7:48
• It does not help us in itself, the theorem rather goes like this: If and only if $(\alpha,\beta) = a_n$ (for a particular number sequence $a_n$), we can construct a Steiner chain with $n$ circles. The reason being that, after inverting into two concentric circles with $e^{a_n}$ as their radius, these admit a Steiner chain (consisting of $n$ congruent circles, of course). One can just invert this chain again to obtain a chain for the original pair. – Josef E. Greilhuber Aug 20 '19 at 6:02