I want to show that for all $n$ there is some collection of $n + 2$ circles such that two of the circles ($A$ and $B$) are tangential to each of the remaining $n$ circles (but not to each other) and each of the $n$ circles are tangential to both $A$ and $B$ but not to any of the other $n$ circles. My initial conception was to let $A$ and $B$ be two large circles (as large as necessary) that are extremely close to each other but not tangential. Then we draw one circle beneath $A$ and $B$ that is tangent to both of them. Slightly above it and in between the "AB crevice" we draw another circle. Then slightly above that we draw another. And so on and so forth until all the circles have been drawn. It seems to me in an intuitive sense that we could draw an infinite number of such circles, but I can$t seem to find a rigorous mathematical expression of this idea.

Note that if this idea is simply wrong, or if you have a simpler way to approach this problem, those would be acceptable answers as well.

  • 4
    $\begingroup$ Isn't it simpler to have two concentric circles of close but distinct diameters and a lot of circles in between? $\endgroup$
    – tomasz
    Commented Dec 4, 2013 at 4:43
  • $\begingroup$ Maybe - honestly I'm not sure how to prove that there's an infinite number of circles you could draw in that case either. $\endgroup$
    – Kvass
    Commented Dec 4, 2013 at 5:06
  • $\begingroup$ I was under the impression that you wanted a given finite number of circles and not an infinite number? You can't get that in this manner. $\endgroup$
    – tomasz
    Commented Dec 4, 2013 at 5:11
  • $\begingroup$ Yes, sorry, a finite number n, but where n can be any value. $\endgroup$
    – Kvass
    Commented Dec 4, 2013 at 5:37
  • $\begingroup$ You might be interested in Steiner Chains. In such a construct, the $n$ circles other than $A$ and $B$ make a tangential chain, which isn't strictly what you want, but that's easy to remedy: Make a Steiner Chain with $2n$ circles, and simply erase every other one. (I'll note that Steiner chains are usually depicted with circle $A$ inside circle $B$, but inversion can put them outside each other.) $\endgroup$
    – Blue
    Commented Dec 4, 2013 at 7:15

3 Answers 3


You could take the vertices of a $m$-agon width unit length sides as the centers of $m$ circles with radius equal to $\frac{1}{2}$, then take $A$ and $B$ as two concentric circles, with center in the center of the $m$-agon, having radii $\frac{1}{2\sin\frac{\pi}{m}}\pm\frac{1}{2}$. As pointed by @zyx, circular inversion brings this configuration into any configuration you may like more, like the one in which $A$ and $B$ are external circles with equal radii: just invert with respect to a circle centered on the incircle of the $m$-agon.

The following picture shows the $m=5$ case, for instance: $A$ and $B$ are the orange circles.

A Steiner chain

As pointed by @Blue, you can simply carry on this construction for $m=2n$ then erase the even-indexed red circles, or just invert a chain of $n$ non-touching circles inscribed in a circular annulus, as depicted below ($n=5$).

A modified Steiner chain

  • $\begingroup$ Just to make sure I understand - the black circles are before the inversion, and the red (and orange) circles are after the inversion? $\endgroup$ Commented Mar 8, 2014 at 18:06
  • $\begingroup$ @ErelSegalHalevi: yep ;) $\endgroup$ Commented Mar 9, 2014 at 15:07

Can you find a collection in which the two large circles (A and B) are also disjoint (not overlapping)?

Yes. "Not overlapping" apparently means that interiors should be disjoint.

Ignoring at first the non-overlap condition, the only position of the circles where there is an easy way to construct many additional tangent circles is when $A$ and $B$ are concentric circles of an annulus (thin enough relative to its radii that more than $n$ tangent circles fit in the ring).

Inversion (http://en.wikipedia.org/wiki/Circle_inversion) in a point not on any of the circles transforms the circles to circles, preserving all relationships of tangency, non-tangency, intersection and non-intersection. The center and radius of the inversion can be chosen to arrange that the interiors of the transformed versions of $A$ and $B$ and (I think) the other circles, are disjoint.

  • $\begingroup$ It doesn't matter if the small circle intersects the rest of the diagram, because inversions take circles to circles. $\endgroup$
    – Carl
    Commented Mar 6, 2014 at 0:44
  • $\begingroup$ Can you give a simple sketch of how it looks like after the inversion? $\endgroup$ Commented Mar 6, 2014 at 10:57
  • $\begingroup$ $A$ and $B$ are required not to be tangent to each other in the original question. That's not too hard to arrange, but inverting parallel lines will not do. $\endgroup$
    – user856
    Commented Mar 6, 2014 at 11:13
  • $\begingroup$ Thanks [and the answer updated accordingly]. It means that the solution is essentially unique, in that one has to start with concentric circles. @Rahul $\endgroup$
    – zyx
    Commented Mar 6, 2014 at 17:48
  • $\begingroup$ @ErelSegalHalevi, I'm not sure how to make diagrams, but to draw the picture, take two external circles A and B, then start drawing externally tangent circles between them by first drawing one (C), then another (D) externally tangent to A B and C, then another E tangent to A B and D, then F tangent to A B and E, etc. As in Blue's comment, A and B and the odd-numbered circles one has just drawn (C and E and G and ...) are a particular example of what the end result could look like. $\endgroup$
    – zyx
    Commented Mar 6, 2014 at 17:55

For the $k^{th}$ small circle where $1 \le k \le n$, we first center them along the $x$-axis at $(k-1,0)$ with some radius $r_k$ to be determined. Now pick a $R$ so large such that $\sqrt{R^2 + n^2} \le R + \frac16$. We then center the two big circle at $(0,R)$ and $(0,-R)$ with radius $R -\frac13$. It is clear $A$ and $B$ are disjoint from each other. In order for the $k^{th}$ small circle to tangent to both $A$ and $B$, $r_k$ is given by

$$\begin{align} & r_k = \sqrt{R^2 + k^2} - \left(R-\frac13\right)\\ \implies & r_k < \sqrt{R^2 + n^2} - \left(R-\frac13\right) \le \left(R +\frac16\right) - \left(R-\frac13\right) = \frac12 \end{align}$$

This means the $n$ small circles are also disjoint from each other and yet tangent to both $A$ and $B$.


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