Eight circles touching three

Three circles, drawn anywhere so long as none is inside another. A fourth circle that touches all three: how many are there?

The answer is eight, and the reason is a counting argument with no geometry in it. Touching a circle means one of two things — lying outside it, or containing it — so a circle tangent to three others has made three independent binary choices. Two to the power of three is eight, and for three circles in general position each of the eight choices is realised exactly once.

The equation, and why it is not one equation #

Write the unknown circle as a centre $c$ and a radius $r$. Tangency to a given circle $(c_k, r_k)$ is

$$|c - c_k| = r + r_k \quad \text{or} \quad |c - c_k| = |r - r_k|,$$

according to whether the two lie outside each other or one contains the other. Writing $s_k = \pm 1$ collapses both into $|c - c_k| = r + s_k r_k$, and the sign is the choice.

Three of those with the signs fixed is three equations in three unknowns, and they are not linear. But subtracting one from another cancels both $|c|^2$ and $r^2$, leaving something linear in the centre’s coordinates and the radius. Two subtractions give two linear equations; solving them writes the centre as a straight line in $r$; substituting back into any one original equation leaves a quadratic in $r$ alone.

So each sign pattern costs one linear solve and one quadratic, and the whole problem is eight of those. That is what the figure does, and it explains a feature of the picture: the eight circles vary enormously in size, because the sign pattern that puts the unknown circle outside all three and the one that makes it contain all three are asking for very different objects.

Every root is checked against all three original tangency conditions before it is drawn. An elimination that has cancelled two squared terms can produce a solution of a different equation, and the check costs nothing.

Where the count can fail #

The eight is a theorem with hypotheses, and the hypotheses are geometric rather than algebraic — which is a good sign that the count is about the configuration rather than about the method used to find it.

Eight is the answer in general position, and the failures are worth knowing because they are the cases a construction has to handle.

If the three centres are collinear, the two linear equations become dependent and the elimination has nothing to solve. If two of the given circles are tangent to each other, some of the eight solutions coincide. If one circle is inside another, some sign patterns have no real solution at all and the quadratic’s discriminant goes negative. And three circles through a common point admit a whole family of tangent circles rather than eight.

The figure refuses the degenerate inputs rather than drawing a partial answer: it checks that the three given circles lie outside one another and that their centres are not collinear, and complains otherwise. That is the honest response to a count that depends on a hypothesis — a picture of six circles labelled “eight” would be worse than no picture.

It is worth being explicit about the cost of the elimination, because it is the reason a two-thousand-year-old geometric method is still the one worth knowing.

Writing the three tangency equations and eliminating gives, after the dust settles, a quadratic whose coefficients are polynomials of degree four in the six input numbers. Solved symbolically the expressions run to pages. Solved numerically — which is what the figure does — they are fine, and they are fine only because every root is checked afterwards against the conditions it was supposed to satisfy.

Without that check the method is untrustworthy in a specific way. Subtracting two tangency equations produces a consequence of them, not an equivalent of them, so a solution of the eliminated system need not solve the original; the extra roots are solutions with the wrong sign convention, and they look exactly like the right ones. The check separates them in one line, and it is the only part of the computation that could not be omitted.

That pattern — eliminate, solve, verify — is the shape of every algebraic attack on a geometric problem, and the third step is the one that is skipped. The same discipline is what makes a construction with straightedge and compass a proof rather than a drawing: the steps are exact, and what makes them trustworthy is that each one’s output satisfies a stated condition rather than merely looking right.

The classical solution, which is inversion #

The elimination above is what a computer does. It gives no insight, it does not generalise, and it is not how the problem was solved for two thousand years.

Apollonius’s approach, in the form Viète gave it, is a reduction. The key observation is that the problem gets easier when one of the given circles is a point, because a circle through a point can be inverted at that point and becomes a line.

Step one: shrink. Increase or decrease all four radii — the three given ones and the unknown — by the same amount. Tangency is preserved: if $|c - c_k| = r + r_k$ then $|c - c_k| = (r - t) + (r_k + t)$ for any $t$. Choose $t$ to be the smallest given radius, and that circle becomes a point while the other two remain circles.

Step two: invert. Invert the whole configuration at that point. The two remaining circles become two circles, and the unknown circle — which passes through the centre of inversion — becomes a line.

Step three: the easy problem. A line tangent to two circles is a common tangent, and two circles have four of them, found with a ruler.

Step four: come back. Invert the tangent lines back into circles through the point, and un-shrink the radii.

Four common tangents, and the choice of which circle to shrink accounts for the rest of the eight. The whole difficulty was moved into a position where it evaporated, which is what a transformation is for.

What the reduction is really doing #

The pattern is worth isolating because it is the reason inversion earns a place in a geometer’s toolkit rather than a formulary.

A transformation is useful when it preserves the hypotheses of a family of problems and simplifies their conclusions. Inversion preserves two things and almost nothing else: the class of circles-and-lines, and angles. A tangency problem is stated entirely in those terms — tangency is the angle between two circles being zero — so every tangency problem is carried to another tangency problem.

That is the whole licence. It means one may choose the position freely, and the right position is chosen by looking at what would make the conclusion obvious. Here it was make one circle a point, so that its image is a line. In Steiner’s porism it was make two circles concentric, after which the closing of a chain is a division of a full turn. The habit generalises past circles: choosing coordinates in which a conic is a standard one is the same move made with a different group.

What the eight look like #

Reading the picture rather than the count is worth a paragraph, because the eight are not eight of the same thing.

One of them lies outside all three given circles and is the smallest — it nestles in the gap between them. One contains all three and is the largest. The other six are mixtures, each containing some and avoiding others, and they come in three pairs related by which single circle is treated differently from the other two.

That structure has a name in the classical literature: the eight solutions form four pairs, each pair consisting of a circle and the one obtained by swapping every choice. The two members of a pair are the two roots of the same quadratic, and their radii are related by the coefficients — which is why the largest and the smallest appear together as the all-outside and all-inside pair.

Nothing in the figure marks the pairing, and a reader can find it by looking: the circle in the gap and the circle round everything are the extremes of the same family.

Counting before constructing #

There is a habit visible in this problem that is worth naming, because it recurs throughout the subject.

The count came first, and it came from the statement rather than from any attempt at a construction. Three binary choices means eight, before anyone knows how to find one. That kind of argument is cheap, it is often available, and it tells a constructor when to stop looking.

It also tells a constructor when their method is incomplete. A ruler-and-compass procedure that produces four tangent circles has not solved the problem; it has solved half of it, and the count says exactly how much is missing. The same accounting decides how many regions a set of nearest-neighbour boundaries cuts a plane into and how many spanning trees a graph has — a number derived from the description, checked afterwards against a construction that produces them one at a time.

Special cases, and how many there are #

The problem has a family of degenerate versions and they were classified long before the general one was solved, because each is easier and each was met first.

Replace a circle by a point and tangency to it becomes passage through it. Replace a circle by a line and tangency to it becomes touching. With three objects each of which may be a point, a line or a circle, there are ten cases up to symmetry, and they run from three points — one circle, the circumscribed one, and the classical construction is Euclid’s — to three circles, which is this one.

The ten cases are not ten problems. Each is the general one with some of the eight solutions coincident or at infinity: three points give one circle because seven of the eight have collapsed together, three lines give four because the choices reduce, and so on down the list. Seeing them that way is the modern reading and it took the general solution to make it available.

Where the problem comes from and where it goes #

Apollonius of Perga posed and solved this around 200 BC in a lost work; what survives is Pappus’s account of it. Viète reconstructed the solution in 1600, and the sequence of reductions above is essentially his. Newton gave a construction using conics, Gergonne a projective one in 1814, and the inversive solution — the one that makes the reduction obvious — had to wait for inversion itself, in the 1820s.

That is a long time for a problem whose answer is eight. The reason is that the reduction needs a transformation nobody had, and the transformation was not discovered by anyone trying to solve this.

The subject continues in two directions. Upwards in dimension: spheres tangent to four given spheres, sixteen of them, by the same sign count and the same reduction — and the arithmetic that decides how many spheres fit in a box is a different subject that shares the vocabulary and none of the method. Inwards: apply the construction repeatedly and the circles fill the gaps forever, which is the next rung — and the curvatures of the circles that appear turn out to be whole numbers.

What the pictures cannot show #

The construction is described and the solutions are computed. Nothing on this page draws the shrink-and-invert reduction being carried out; the eight circles were found algebraically and checked geometrically, and the classical construction is prose.

Degenerate configurations are refused rather than drawn. Collinear centres, nested circles, circles through a common point — each breaks the count in its own way, and each is a picture this figure will not produce.

The eight are not all visible at once at a useful size. The largest of them dwarfs the given circles, so a canvas holding all eight shows the three given circles small. That is a fact about the answer rather than about the drawing.

And nothing here shows the count failing. A figure of a configuration with six tangent circles rather than eight would be the useful complement to this one, and the degeneracies that produce it are exactly the inputs the generator refuses.

Where the ladder goes next #

The rung above takes the construction and iterates it. Given three mutually tangent circles there are two circles tangent to all three; take one, and now there are more triples, each with its own two. Repeating forever fills every gap, and the resulting packing has a property nobody would predict from the construction: if the first four curvatures are whole numbers, every curvature in the infinite packing is a whole number.

The relation that makes that true is Descartes’s, and it is a quadratic whose two roots sum to twice something — so the second solution is obtained from the first by subtraction, with no square root anywhere.

What is worth carrying away #

A count and a construction answer different questions, and getting the count first changes what a construction has to do.

Eight comes from three binary choices and is available before any geometry. Every method for actually finding the circles is then measured against it: a method producing four has found half of them, and a method producing one has found a special case. The count is a specification, and the construction is an implementation — and it is much easier to notice that an implementation is incomplete when the specification was written first.