Today I want to talk about a geometry I think is grossly overlooked, especially when compared to the popularity of its Euclidean brother. Today I want to talk about inversive geometry. Inversive geometry takes the standard plane we know and quite literally flips it inside out. By the end of this post, you will be familiar with not only what the heck an inversion is, but a very cool theorem that results in the animation above that relates tangent circles to one another. But before we can get there, we first need to learn how to flip our world inside out.

Plane Inversions

As you can imagine, inversive geometry is geometry that relies on something called inversions. You can think of an inversion as a function that takes a point $P$ and spits out a transformed point $P'$. But, what exactly is our function? It's not a standard $f(x)$ as we're giving two coordinates not one. So maybe it's a 2-by-2 matrix, as we're giving a 2D vector and outputting another 2D vector? Not a bad idea, but it will quickly become clear why we don't want to do that. So, what is our functional object? It's actually a circle. As weird as that sounds, just hear it out. Given a circle $Ø$ with center $O$ and radius $r$, the point $P$ is inverted to $P'$ based on the following equation: $OP \cdot OP' = r^2$ where $P'$ lies on the ray $\overrightarrow{OP}$. Try dragging the points below to get a handle on this idea.

The numbers above each point represent their distance from $O$, so you can verify the distances satisfy the inverting equation. Just within a few seconds, though, I'm sure you can see why this is called an inversion. Every point on the inside of the circle gets mapped to the outside of the circle, and every point outside the circle, gets mapped to the inside (and every point on the circle stays on the circle—we say the circle is invariant under inversion). We're taking our plane and flipping it inside out centered around the circle. As such, this is specific inversion is known as a circle inversion or plane inversion. It should now be apparent why matrices won't work as this not only doesn't preserve the origin, but it's also not a linear transformation (the scaling of $P'$ relative to $P$ is directly proportional to $P$ itself). One thing worth noting is that we have to formally introduce a point at infinity. That way, if we try to invert the center of our inverting circle (i.e. $O$ in the above circle), we have a place for it to go to. Now that we can invert points, we can now easily invert shapes. All we have to do is invert the collection of points individually, and remember the order to connect them. We could try basic polygons like squares and triangles, but the one that is most interesting (and will be most helpful) is inverting other circles. Below, we'll again invert over the green circle with center $O$, but now instead of a point, we'll invert the blue circle with center $C$ to the red circle.

A lot of our rules with inverting points can easily give us an intuition for how our circles might invert. Points on a circle inside of our circle of inversion get flipped to be outside of it and vice versa, and points on the circle of inversion stay on the circle of inversion. But, since we are looking at a group of points, some discrepancies between points obviously exist. For instance: distance. I have drawn both the center of our blue circle $C$ as well as its inversion $C'$. But just by looking at it, it's obvious that there's no way $C'$ can be the center of the blue circle's inversion! That's due to one key aspect of inversions: they do not preserve distances. That should be apparent due to the actual inversion equation $OP \cdot OP' = r^2$. This inverse relationship between the length of $OP$ and $OP'$ is what really exaggerates inversions with very small or very big values of $OP$. In fact, this is why inverting a circle is so interesting as even though the distances get all messed up, a circle will still always invert into another circle. Try inverting a square and you'll almost definitely get something that doesn't look like a square (almost definitely as if you align the center of the square with the center of the inversion circle, then that will result in another square; just try drawing it out and it will all fall into place). Even when it doesn't look like a circle, it really is! Try dragging the blue circle such that it intersects the center of our green inverting circle. You'll get something that looks like a line. While it acts as a line, we formally say that this is a really really big circle. Specifically, a circle with an infinitely large radius. It's a lot like how in calculus they say if you zoom in super far in on a curve it looks like a line, if you have a super big curve it locally looks like a line from our perspective.

Ok, so what does this tell us? Well, just inverting the one circle tells us a lot about inversions and some properties they have. Let's write them out.

  1. Inversions do not preserve distances. We saw this with how a circle's center may not invert to the center of the inverted circle.
  2. Every point $P$ has a unique inversion $P'$ for any given circle of inversion $Ø$ with center $O$ and radius $r$. This may seem obvious, but it's important to be aware of as it leads to the next very important characteristic...
  3. Intersections and tangency between two or more shapes are preserved during an inversion. This fact is the one you want to hold onto the most for the upcoming sections. This should make sense as if two or more shapes share some point in common such as an intersection or tangent point $P$, that singular point has only one unique inverse $P'$ which they must also share. And if all the points need to be connected to that point after the inversion, then we should expect to see that intersection/tangent point remain after an inversion.
  4. Lastly, just as a neat fact, performing the same inversion twice results in nothing changing (the identity). You can think of this sort of like what happens when turning a shirt inside out twice: the first time the seams come out, but the second time it just goes back to how it started. Just going back to our formula $OP \cdot OP' = r^2$, we know that $OP'$ has some length that corresponds with $OP$ to keep that product the same. So, if we let $P \rightarrow P'$ representing our first inversion, $P'$ needs to go back to $P$ to keep the formula the same for the second inversion. This makes the function of an inversion a special function called an involution.

With the basics of inversion down, we are now ready to explore that animation from the very top of this post.

Steiner's Porism

Steiner's Porism can sound a lot more complicated than it is, but I promise you the animation at the top explains everything. First, draw two circles with one inside the other. Second, draw a third circle that just touches the inside and outside circle. Third, draw as many circles as you can fit touching the inside, outside, and last circle you drew until no more can fit. Steiner's Porism states if the last circle touches the first circle you drew, then no matter where you draw the first circle, the chain will always connect (and if the last circle doesn't touch the first, then they will never touch). So in the GIF above, the black circles are free to rotate like ball bearings in between the outer blue and inner red bounding circles and the chain will always link up. It's a pretty neat fact, but how would someone ever prove this? That's where our friend inversion comes in.

Before we go any further though, let's quickly see if there's an easier version of this scenario. One of the main issues I first had when looking at this was the fact that the black circles rolling around didn't have to be the same size. Fortunately, there's an obvious case where we don't have to worry about that: when the two bounding circles are concentric.

When the circles share the same center, then Steiner's Porism becomes obvious: our set up now becomes symmetrical, so you can think of any starting point as a rotation of the original chain of circles. Since this is an easy fact to see, we can now use a key property of inversions to prove the general case for any pair of bounding circles:

Intersections and tangency between two or more shapes are preserved during an inversion

This is great for us, as then if we can find an inversion that turns our non-concentric circles into concentric circles, we can then use the fact that our tangents of the chain circles are preserved and use the obvious concentric circles case to close out the theorem. It may sound complicated, but think of it as a way to work backwards: if we can show we can turn any non-concentric circles into concentric ones, then the reverse is also true where there is some corresponding pair of concentric circles that invert to our original, non-concentric ones. Since the rules of tangency remain true between inversions, the rules for our circle chains remain as well (since they are only governed by tangents).

To find our desired circle of inversion, I'll present it as a series of steps that might not make sense immediately, but will definitely make sense retroactively. So, for now, I ask that you follow along with the steps and we'll discuss it at the end.

Step 1: Find the radical axis

The radical axis of a pair of circles is the line (or axis, I guess) that every point $P$ along the line is the same distance away from the tangents of the two circles. Fortunately, it's not too hard to find with some simple geometry. We'll draw the radical axis in green.

Of course, the point $P$ in question has to be outside the two circles to be able to find tangents, but that's only a worry for intersecting circles (which we don't care about). I drew a purple circle center around $P$ to show that the tangents are in fact equal in length, but this purple circle has one notable property: it is orthogonal to the other 2 circles, meaning that it intersects the green and blue circles at right angles. This fact is a result that a circle's radius is perpendicular to its tangent. Speaking of orthogonal circles, that brings us to our next step.

Step 2: Draw two orthogonal circles and find their intersection

This step is easy enough since we've basically already done half of it. We just need to draw another purple orthogonal circle as we've done before, and then find their intersection. Our space will start to get cluttered quickly, so I'll remove the purple tangency lines, but just know that those are what determine our purple circles. We'll draw this intersection point in black.

Here I've selected the outermost intersection for clarity, but we'll see in just a second that either of the two intersections work just fine. First, it's worth noting that for a given configuration of outer blue and inner red bounding circles, the intersection points remain constant. No matter how you may slide those green points, the intersection point doesn't change. That should help cue you into its importance. As a separate, interesting fact (that I haven't looked into enough), the centers of the red and blue circles are collinear with the two intersection points of the purple circles. But with that out of the way, we can move onto the third and final step.

Step 3: Draw any circle at the chosen intersection point

Notice how I said "any" circle. A circle with any radius will suffice as our desired circle. This will be our circle to invert over! We're going to invert a total of four circles: of course, the red and blue ones, but we'll also invert the two helper, purple circles.

And just like that, we've obtained our concentric circles just as we desired! Just to reiterate, because tangencies are preserved through our inversion, we can then draw our chain of circles in the original blue and red bounding circles and know for a fact that they'll remain tangent after our inversion as well. Moreover, since concentric circles present a symmetric case, Steiner's Porism is nicely proved. One thing worth noting, though, is that we get a solution even when the two circles are not contained within one another. If the two circles are non-intersecting and completely separated from one another, we can still follow our procedure from before: we can find a radical axis of the two circles, which leads to our two orthogonal circles, that finally intersect at the center of our inversion circle. However, we now get a reversed solution with the red circle becoming the outer concentric circle instead of the inner one (this only happens as a result of the choice of intersection point of our orthogonal circles).

But, why does this even work? I mean, yeah, it produces concentric circles, but why is that? It has to do with our purple orthogonal circles. Remember, these circles are orthogonal meaning that they intersect our blue and red circles at right angles. Moreover, remember that by definition of our construction in Step 2, these orthogonal circles pass through the center of our black inversion circle, and as we saw from before, that means that these purple circles will invert into circles of infinitely-large radii (or lines, if you prefer—you can actually see the "lines" phase in and out at certain angles due to the approximation of this program). Lastly, we also know that intersections are preserved during an inversion. So, not only do we know that our inverted purple lines must intersect, but the intersections between the red and blue circles as well as the purple circles are also maintained. So, we have two lines that intersect that need to be orthogonal to two other circles. What configuration allows this? The only way the lines can be orthogonal to the circles after inversion is if those lines are the radial lines of the circles! So, both circles must share the same center (the intersection of the lines) such that the lines become radial, and by definition of sharing a center, they must be concentric!

Inversive geometry has all sorts of interesting quirks and facts to explore, and should be more well known than what it is. Maybe one day I'll touch on its connection to polar curves. But anyways, this post wouldn't be complete if you couldn't build a Steiner chain of circles of your own, so below there is one last widget to experiment with tangent circles. I have left the special black inversion circle on the canvas just so you can see how all of our work to get concentric circles relates to any pair of nested, bounding circles. There's so much I had to gloss over to keep this short, such as the hidden conics in the path of the tangent circles, so I highly recommend skimming other articles such as Wolfram MathWorld's and even Wikipedia's discussions on Steiner chains. With all things in math, this story is never over: Steiner's Porism has a projective geometry cousin known as Poncelet's Porism, but that deserves its own post entirely some day. With that, enjoy the little animation below. As before, don't forget to try separating the circles to get some strange, but special solutions to Steiner's Porism (if you're having trouble seeing the animation clearly, try reducing the radius of the black inversion circle).