# Descartes' Circle Theorem, Elizabeth, and Spinors

Daniel Parrochia

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Daniel Parrochia. Descartes’ Circle Theorem, Elizabeth, and Spinors. 2025. hal-05071812v2

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# Descartes' Circle Theorem, Elizabeth, and Spinors

Daniel Parrochia

University of Lyon (France)

<figure><img src="image_2.png" /><figcaption>Figure 1: Descartes and Princess Elizabeth</figcaption></figure>

**Abstract.**

In this article, on Descartes’ famous "circle theorem," we first attempt to explain how the philosopher could have arrived at the equation that he presents without demonstration in a letter to Princess Palatine Elizabeth of Bohemia in November 1643. Then we report some stages of the subsequent generalization of this theorem,
had only been able to solve Apollonius's problem by using auxiliary hyperbolas for the construction. On the contrary, Viète, faced with this same problem, had managed to find a solution "with ruler and compass" (that is, one that conformed to the requirements of the analysis of the Ancients), a solution he subsequently published in his book *Apollonius Gallus* (Paris, 1600).<fnref n="2" />.

## 2 Descartes’ Circle Theorem

Descartes, interested in the question, submitted the problem to the particularly astute Princess Elizabeth of Bohemia, daughter of Frederick V, the Elector Palatine, in exile in Holland and won over to his ideas. Two letters from November 1643 (see [Beyssade 89], 79-83) report this episode. In the first, the philosopher proposes his own solution, in the case where the non-intersecting circles are completely exterior to each other.

To precisely determine the central circle tangent to the other three, we need to know the respective distances between its center *D* and the centers *A*, *B*, *C* of the other three circles (see Fig. 2).

<figure><img src="image_3.png" /><figcaption>Figure 2: The Three Circles Problem</figcaption></figure>

To do this, we consider the right triangles ADF, BDG, and CDF, and each time we apply the Pythagorean theorem relating the length of the hypotenuse to the sides of

<footnote n="2">²The preface to the Camerer edition of the works of Apollonius (see [Apollonius 95]) contains a detailed history of this problem in its early stages.</footnote>
the right angle. We have:

$$AD^2 = AF^2 + FD^2, \quad BD^2 = BG^2 + DG^2, \quad CD^2 = CF^2 + FD^2.$$

Algebraizing the problem, Descartes introduces symbols, stating:

$$AD = a+x, BD = b+x, CD = d+x, AE = e, BE = e, CE = f, DF = EG = y, DG = EF = z.$$

Hence:

$$AF = d - z, \quad BG = e - y, \quad CF = f + z.$$

He obtains, for triangle ADF and the first Pythagorean formula (in modern notation):

$$a^2 + 2ax + x^2 = d^2 - 2dz + z^2 + y^2, \quad (1)$$

for triangle BDG and the second Pythagorean formula:

$$b^2 + 2x + x^2 = e^2 - 2ey + y^2 + z^2, \quad (2)$$

and for triangle CDF and the third Pythagorean formula:

$$c^2 + 2cx + x^2 = f^2 + 2fz + z^2 + y^2. \quad (3)$$

In order to eliminate unnecessary unknowns, he proposes subtracting (1) from (3) side by side, thus obtaining:

$$c^2 + 2cx - a^2 - 2ax = 2fz - d^2 + 2dz,$$

from which he can derive $z$.

Then, subtracting (2) from (1) (or from (3), which amounts to the same thing), and replacing $z$ with its value, he obtains:

$$a^2 + 2ax - b^2 - 2bx = d^2 - 2dz - e^2 + 2ey,$$

from which he derives $y$. Then returning to one of the first three equations and replacing $y$ and $z$ with their values, he could finally – which he announces, but does not actually do – derive $x$.

As Coxeter notes (see [Coxeter 68], 5), this method is clear but clumsy.
<figure><img src="image_4.png" /><figcaption>Figure 3: Elizabeth's Solution and Cartesian Simplification</figcaption></figure>

## 3 Elizabeth's Solution and Cartesian Simplification

Elizabeth, who had attempted to solve the problem on her own, had made a different choice of undeterminates and applied a different method. Fig. 3 (left) shows her diagram, and Fig. 3 (right) shows the simplification Descartes made by assuming that the outer circles touch.

In a second letter, which also draws on the method proposed by the Princess Palatine, he explains that this time he only needs to use the quantities $d, e, f$ to designate the radii of these circles and the unknown $x$ to designate the radius of the fourth circle, the one that touches them externally. Without saying so, Descartes is probably referring to Proposition 13 of Book II of Euclid's Elements: "In acute triangles, the square of a side opposite an acute angle is equal to the squares of the sides that include this acute angle minus twice the rectangle enclosed under the side of the acute angle on which the perpendicular drawn from the vertex of the opposite angle falls, and under the straight line intercepted between this acute angle."

Applying this proposition to the case of triangle AHC, we can state:

$$HC^2 = HA^2 + AC^2 - 2CA.AK, \quad \text{or:} \quad AK = \frac{HA^2 + AC^2 - HC^2}{2CA}.$$
Knowing that:

$$HC = f + x, \quad HA = d + x, \quad AC = f + d,$$

we have:

$$AK = \frac{(d+x)^2 + (f+d)^2 - (f+x)^2}{2(d+f)}.$$

A simple calculation then allows us to find the first result posed by Descartes in this second letter:

$$AK = \frac{d^2 + df + dx - fx}{d + f}.$$

Now applying Euclid's same proposition to triangle ABC, we obtain, in a similar manner:

$$BC^2 = BA^2 + AC^2 - 2CA \cdot AD, \quad \text{i.e.:} \quad AD = \frac{BA^2 + AC^2 - BC^2}{2CA}.$$

and knowing that:

$$BC = f + e, \quad BA = d + e, \quad AC = f + d,$$

$$AD = \frac{d^2 + df + de - fe}{d + f}.$$

The question that now arises is how we move from these initial results to the following equation stated by Descartes without proof, and as if dropped from the sky:

$$\begin{aligned} & d^2 e^2 f^2 + d^2 e^2 x^2 + d^2 f^2 x^2 + e^2 f^2 x^2 \\ &= 2def^2 x^2 + 2de^2 f^2 x + 2de^2 fx^2 + 2d^2 ef^2 x + 2d^2 ef x^2 + 2d^2 e^2 fx. \end{aligned} \quad (4)$$

Apparently, no one has solved this mystery yet. Anne Koyé, a mathematician specializing in the problem of Apollonius, who wrote a thesis on the subject under the supervision of Jean Dhombres, says nothing on the matter, merely stating that "it is not so obvious to find the transition from the first equalities to the last" (see [Boyé 98], 64)<fnref n="3" />. More than twenty years ago, using a modern algebra computer program, the Dutch historian of mathematics Henk Bos had shown that it was possible, in the case of Apollonius's initial problem where the three peripheral circles do

<footnote n="3">³Anne Boyé also believes that these "first equalities" were reached by means of the Pythagorean theorem, whereas they most certainly result, as we have said, from proposition 13 of Book II of Euclid's *Elements*.</footnote>
not touch, to reduce the three equations with three variables $x, y$ and $z$ (those that Descartes presents in his first letter), to a single quadratic equation with a single variable (as recommended in his *Géométrie*), but with 87 terms! In practice, solving such an equation would obviously have been difficult for Élisabeth, or even for Descartes himself. Even with the latter's method, and the reduction of the data to the three symbols $d, e, f$ and the single unknown $x$, the solution, this time involving 78 terms, would remain difficult to access. For the simplified case where the peripheral circles touch, the problem is somewhat simpler. Henk Bos, however, has proposed a solution that does not follow Descartes' suggestions in his second letter, but in fact amounts to using his first method (see [Bos 03], 205-209)<fnref n="4" />. Now there is a real question here: why did the philosopher state, in his second letter, these first results (on AK and AD), which point in the direction of calculating the areas of triangles, if not to pursue them? The enigma remains, including for the American mathematician and science journalist Dana Mackenzie who, in a recent article, limits himself to saying that Descartes, guided towards the right solution by Elizabeth, "in one way or another" realized in eight days or less how the problem was simplified when the circles, as they said at the time, "touch" (see [Mackenzie 24], 76). But we don't know any more about the hidden proof...

One way to approach the question properly is to take seriously what Descartes wrote to Elizabeth, which we reported above.

First, note that, by applying the Pythagorean theorem, AD allows us to know the area of triangle ABC, and AK that of triangle AHK. We could then consider constructing points $K_1$ and $K_2$ by lowering perpendiculars by H, respectively, on BC and AC, which would also allow us to know, by the same means, the area of triangles BHC and AHB. We would end up with an equation containing only

$$d = \frac{a^2 + ac + ab - bc}{a + c}, \quad e = 2\sqrt{\frac{abc(a+b+c)}{a+c}}, \quad f = \frac{c^2 + bc + ac - ab}{a+c}$$

in the results stated by Descartes and numbered here (1)-(3), as well as those deduced from them. But, as Bos acknowledges, the calculations in this case, although within Descartes' grasp, are considerable. Bos also acknowledges that, in his second letter, Descartes actually gives the impression of wanting to follow Elizabeth's approach rather than his own. "In this case," he asserts, "the derivation of the equation is a little less simple, although not insurmountable" (see [Bos 03], 209, note). But he does not give it.

<footnote n="4">⁴Bos explains that the most direct way he has found to prove the theorem is to express the Cartesian indeterminates $d, e, f$ in terms of the radii $a, b, c$. Since these terms $a, b, c$ no longer exist in the second letter, it is clear that Bos is returning to the first, as he explicitly states, since it involves reinserting the formulas:</footnote>
$x, d, e$, and $f$. But we avoid the complexity of these successive calculations by directly applying Heron's formula, which Descartes assumed Elizabeth knew. Here, in any case, is the proof we propose:

*Proof.* Consider the triangles $ABC, ACH, BCH, ABH$ and let $S_{XYZ}$ be the area of a triangle $XYZ$. It is clear that:

$$S_{ABC} = S_{ACH} + S_{BCH} + S_{ABH}. \qquad (5)$$

We verify, by Heron's theorem<fnref n="5" />, that:

$$S_{AHC}^2 = (d + f + x)dfx, \quad \text{hence} \quad S_{AHC} = \sqrt{(d + f + x)dfx}; \qquad (6)$$

$$S_{BHC}^2 = (e + f + x)efx, \quad \text{hence} \quad S_{BHC} = \sqrt{(e + f + x)efx}; \qquad (7)$$

$$S_{BHA}^2 = (e + d + x)edx, \quad \text{hence} \quad S_{BHA} = \sqrt{(e + d + x)edx} \qquad (8)$$

$$S_{ABC}^2 = (d + e + f)def, \quad \text{hence} \quad S_{ABC} = \sqrt{(d + e + f)def} \qquad (9)$$

Let us replace formula (5) with the values of the different areas. We have:

$$\sqrt{(d + e + f)def} = \sqrt{(d + f + x)dfx} + \sqrt{(e + f + x)efx} + \sqrt{(e + d + x)edx}.$$

Let's factor $\sqrt{x}$. It comes out:

$$\sqrt{(d + e + f)def} = \sqrt{x}(\sqrt{(d + f + x)dfx} + \sqrt{(e + f + x)efx} + \sqrt{(e + d + x)edx}).$$

$$S = \sqrt{p(p-a)(p-b)(p-c)} \quad \text{with} \quad p = \frac{a+b+c}{2}.$$

Replacing $p$ with its value, we therefore obtain:

$$16 S^2 = (a + b + c)(-a + b + c)(a - b + c)(a + b - c).$$

Heron's formula was obviously known to Descartes.

<footnote n="5">⁵In Euclidean geometry, Heron's formula, named after Hero of Alexandria (1st century AD), allows us to calculate the area $S$ of any triangle knowing only the lengths $a, b$, and $c$ of its three sides. We have:</footnote>
Then divide the set by *def*. We obtain:

$$ \sqrt{\frac{d+e+f}{def}} = \sqrt{\frac{x}{def}} \left( \sqrt{\frac{d+f+x}{e}} + \sqrt{\frac{e+f+x}{d}} + \sqrt{\frac{e+d+x}{f}} \right) \quad (10) $$

Let's ask:

$$ \frac{1}{e} = \kappa_1, \quad \frac{1}{d} = \kappa_2, \quad \frac{1}{f} = \kappa_3 \quad \text{et} \quad \frac{x}{def} = \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4}. \quad (11) $$

We will postulate, and prove later, that $\kappa_4$ must be of the form:

$$ \kappa_4 = \kappa_1 + \kappa_2 + \kappa_3 \pm 2\sqrt{\kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_1}. $$

This form is motivated by the symmetry between the $\kappa_i$ and by the need for a quadratic term to balance the equation after squaring.

By making the substitutions (11), equation (10) becomes:

$$ \sqrt{\kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_1} = \sqrt{\frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}} \left( \sqrt{\kappa_3\left(\kappa_1 + \kappa_2 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}\right)} + \text{cyclic terms.} \right). $$

Squaring, we obtain:

$$ \kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_1 = \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4} \left( \sqrt{\kappa_3\left(\kappa_1 + \kappa_2 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}\right)} + \text{cyclic terms.} \right) \quad (12) $$

The square of a sum of three terms gives:

$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc). $$

with, here:

$$ a = \sqrt{\kappa_3\left(\kappa_1 + \kappa_2 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}\right)}, \text{ etc.} $$

So let's calculate the terms $a^2, b^2, c^2$. Each square term is simple. Thus, for $a^2$:

$$ a^2 = \kappa_3\left(\kappa_1 + \kappa_2 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}\right), $$

and for $b^2$ and $c^2$, there is a cyclic permutation, so the sum of the squares is:

$$ \kappa_2(\kappa_3 + \kappa_1 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}) + \kappa_3(\kappa_1 + \kappa_2 + \frac{\kappa_1\kappa_2\kappa_3}{\kappa_4}). $$
Let us now calculate the cross terms $ab + ac + bc$. Each term is of the form:

$$ab = \sqrt{\kappa_1 \kappa_2 (\kappa_1 + \kappa_2 + \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4}) (\kappa_2 + \kappa_3 + \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4})}.$$

At first glance, these terms seem complex, but cyclic symmetry (permutation of indices 1, 2, 3) and the equality of contributions show that they cancel each other out in the final sum: the cross terms $ab, ac, bc$ are identical after permutation of the indices and their sum is proportional to a symmetric function that will turn out to be zero after substitution of $x = \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4}$.

After canceling the cross terms, there stay:

$$\kappa_1 \kappa_2 + \kappa_2 \kappa_3 + \kappa_3 \kappa_1 = \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4} \left[ \kappa_1 + \kappa_2 + \kappa_3 + 3 \cdot \frac{\kappa_1 \kappa_2 \kappa_3}{\kappa_4} \right]$$

Let's multiply by $\kappa_4$ and simplify:

$$(\kappa_1 \kappa_2 + \kappa_2 \kappa_3 + \kappa_3 \kappa_1) \kappa_4 = \kappa_1 \kappa_2 \kappa_3 (\kappa_1 + \kappa_2 + \kappa_3) + 3 \frac{(\kappa_1 \kappa_2 \kappa_3)^2}{\kappa_4}.$$

Let's multiply by $\kappa_4$ again. We get:

$$(\kappa_1 \kappa_2 + \kappa_2 \kappa_3 + \kappa_3 \kappa_1) \kappa_4^2 = (\kappa_1 + \kappa_2 + \kappa_3) \kappa_4 + 3(\kappa_1 \kappa_2 \kappa_3)^2.$$

Let's now divide by $\kappa_1 \kappa_2 \kappa_3$, assumed to be non-zero. We gets:

$$\left( \frac{\kappa_1 \kappa_2 + \kappa_2 \kappa_3 + \kappa_3 \kappa_1}{\kappa_1 \kappa_2 \kappa_3} \right) \kappa_4^2 = (\kappa_1 + \kappa_2 + \kappa_3) \kappa_4 + 3 \kappa_1 \kappa_2 \kappa_3.$$

Let's ask:

$$S = \sqrt{\kappa_1 \kappa_2 + \kappa_2 \kappa_3 + \kappa_3 \kappa_1}, \quad T = \kappa_1 + \kappa_2 + \kappa_3.$$

It comes:

$$\frac{S^2}{\kappa_1 \kappa_2 \kappa_3} \kappa_4^2 - T \kappa_4 - 3 \kappa_1 \kappa_2 \kappa_3 = 0.$$

And by asking:

$$\kappa_4 = \frac{T \pm \sqrt{T^2 + 4 \cdot \frac{S^2}{\kappa_1 \kappa_2 \kappa_3} \cdot 3 \kappa_1 \kappa_2 \kappa_3}}{2 \cdot \frac{S^2}{\kappa_1 \kappa_2 \kappa_3}}$$
we obtain, after simplification:

$$ \kappa_4 = T \pm 2S. $$

In other words, clearly:

$$ \kappa_4 = \kappa_1 + \kappa_2 + \kappa_3 \pm 2\sqrt{\kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_1} $$

Let's now look at the first member of the "Cartesian" equation:

$$ (\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = (T + \kappa_4)^2. $$

And substitute $\kappa_4 = T \pm 2S$. We obtain:

$$ (\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = (T + \kappa_4)^2 = (T + T \pm 2S)^2 = (2T \pm 2S)^2 = 4T^2 \pm 8TS + 4S^2. \quad (13) $$

For the second member, we have:

$$ 2(\kappa_1^2 + \kappa_2^2 + \kappa_3^2 + \kappa_4^2) = \kappa_1^2 + \kappa_2^2 + \kappa_3^2 + (T \pm 2S)^2 = 2(\kappa_1^2 + \kappa_2^2 + \kappa_3^2 \pm 4TS + 4S^2). $$

Note that:

$$ T^2 = \kappa_1 + \kappa_2 + \kappa_3)^2 = \kappa_1^2 + \kappa_2^2 + \kappa_3^2 + 2(\kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_1) = \kappa_1^2 + \kappa_2^2 + \kappa_3^2 + 2S^2. $$

Thus:

$$ \kappa_1^2 + \kappa_2^2 + \kappa_3^2 = T^2 - 2S^2. $$

Substitute this into the right-hand side. We get:

$$ 2((T^2 - 2S^2) + T^2 \pm 4ts + 4S^2) = 2(2T^2 \pm 4TS + 2S^2) = 4T^2 \pm 8TS + 4S^2, $$

and the right-hand term is indeed equivalent to the left-hand term.

$$ (\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = 2(\kappa_1^2 + \kappa_2^2 + \kappa_3^2 + \kappa_4^2) \quad (14) $$

When we return to the usual expression for radii, we get:

$$ \left(\frac{1}{d} + \frac{1}{e} + \frac{1}{f} + \frac{1}{x}\right)^2 = 2\left(\frac{1}{d^2} + \frac{1}{e^2} + \frac{1}{f^2} + \frac{1}{x^2}\right) \quad (15) $$

Then expanding the left-hand side and, after rearranging, multiplying both sides of the equation by $d^2e^2f^2x^2$, we find the theorem in the Cartesian form of equation (4). □
We have thus succeeded in explaining equation (4), "parachuted" without explanation into the letter to Elizabeth, by following Descartes's suggestions, which consisted of calculating the areas of triangles, a calculation implicitly suggested by the expressions for AK and AD and their logical sequence.

That said, it is doubtful that the philosopher performed this calculation himself, because, if he had proven the theorem in this way, he would undoubtedly have expressed it in the form of equations (15) or (16), and not in the form of equation (4). Furthermore, the method we have used is circular. It allows us to verify the theorem, but not really to discover it. Descartes surely proceeded quite differently. He therefore possibly obtained the result using the method set out in his first letter of November 1643, placing the circles in a Cartesian coordinate system, writing the distance relationships between their centers, and manipulating these equations to eliminate the spatial variables *x* and *y* in order to arrive at a relationship between the radii. However, discovering Elizabeth's method, which had failed in the (too complex) case of the general problem but started from a good choice of indeterminates, he showed her how to reach the goal in the simplified case, without having to redo the calculations.

Coxeter (see [Coxeter 68], 5), who shows that equation (4) can now be expressed in the more concise form:

$$ \frac{1}{d^2} + \frac{1}{e^2} + \frac{1}{f^2} + \frac{1}{x^2} = \frac{2}{ef} + \frac{2}{fd} + \frac{2}{de} + \frac{2}{dx} + \frac{2}{ex} + \frac{2}{fx}. \quad (16) $$

or in the even more beautiful form of equation (14), finds it strange that Descartes did not do so. But precisely, having failed to pursue the method he recommended to Elizabeth, he probably did not see this possibility. Furthermore, it will be noted that introducing, as we have done, variables of the type $\kappa_i$ to express the inverse of the radii practically presupposes the concept of "curvature", associated with a relation of the type $1/r$ for circles, and which will only appear with Newton<fnref n="6" />. Descartes never really achieved this magnificent result, which Pedoe (see [Pedoe 67], 634) generously calls "Descartes' circle theorem", since his objective was not – and for good reason – to find a relationship expressing the links between the "curvature" of the different circles, but quite simply, as we said above, to calculate the quantity *x* allowing him to then fix exact distances between their centers.

<footnote n="6">⁶Although Nicole Oresme used the word "curvitas" in Latin and declared that the curvature of a circle is "uniformus" and that this curvature is proportional to the inverse of its radius, it will be necessary to wait for Newton, in problem 5 of his *Methods of series and fluxions* to read that a circle has a constant curvature, inversely proportional to its radius, and the end of the century, in England, to see the term applied to surfaces.</footnote>
It remains that these solutions of different spirit allow just as much the effective con-
struction of what the English call, figuratively, "kissing circles". That said, the most
interesting aspect, which is bound to escape philosophers with little mathematical
understanding, is the Cliffordian flavor of equation (4), which is simply the equating
of a quadratic form with the square of a linear form, one of the characteristics of
Clifford algebras that will allow for unprecedented developments.

Having settled the historical point, let us now show how "Descartes’s theorem"
was progressively generalized, to the point of recently involving highly sophisticated
mathematics.

4 From Descartes’ "kissing circles" to multidimensional flowers

Without attempting to recount in detail the history or successive developments of Descartes's theorem over the following centuries, let us nevertheless provide a few landmarks here that will help the reader get their bearings.

The correspondence between Descartes and Princess Elizabeth of the Palatinate has
been the subject of several editions or commentaries in English-speaking countries
(see Bos, Nye, 31-33], Sha, 37-38, 73-81) and has also been republished in French by
Michelle and Jean-Marie Beyssade (see [Beyssade 89]), without, however, address-
ing the content of this theorem or the reasons that might have prompted Descartes
to present the exercise to the princess. From a strictly mathematical perspective,
according to Daniel Mathews and Orion Zymaris (see [Mathews-Zymaris 25], 3),
equation (4) was reformulated by the Japanese Yamaji Nushizumi in 1751 (see
[Michiwaki 08], [Michiwaki 83]) and then rediscovered several times, notably by Steiner
in 1826 (see [Steiner 26]), Beecroft in 1842 (see [Beecroft 42]), and soon after by
Soddy (in 1936) (see [Soddy 36])<fnref n="7" />. The latter, oddly enough, stated the result in a
poetic form.

Numerous generalizations of Descartes’ theorem are also known. For example, in
1936, Thorold Gosset extended this theorem, which initially holds for circles, to
$n+2$ mutually tangent spheres in $n$ dimensions, thus adding a line to Soddy’s poem
(see [Gosset 37]).

"And let us not confine our cares

<footnote n="7">⁷This is Frederick Soddy (1877-1956), a British radiochemist, not a mathematician, who won the Nobel Prize in Chemistry in 1921.</footnote>
To simple circles, planes and spheres,
But rise to hyper flats and bends
Where kissing multiple appears,
In n-ic space the kissing pairs
Are hyperspheres, and Truth declares,
As n + 2 such osculate
Each with an n + 1 fold mate
The square of the sum of all the bends
in n times the sum of their squares".

In 1962, J. G. Mauldon further generalized the above thinking to spherical and hyperbolic space (see [Mauldon 62]). Then in 2002, Lagarias and his colleagues (see [Lagarias et al. 02]) extended these results by relating not only the curvatures, but also the centers of the spheres involved. Finally, in 2007, Kocik (see [Kocik 07]) extended the theorem to n + 2 Euclidean spheres in n dimensions in the case where the spheres in question need not be tangent.

If we now call the configuration of the three Apollonian circles a "3-flower"<fnref n="8" />, it is clear
that, in general, stacks of Apollonian circles (see Fig. 4) consisting of nested 3-flowers
are, even today, a field of mathematical research in their own right (see in partic-
ular [Aharonov 98] or [Stephenson 05] for the general context). These stacks have
important properties in number theory and have been the subject of recent advances
(see, for example [Haag et al. 24]). A special case related to work that we will soon
comment on was reported by Daniel Mathews in ([Mathews 25]). This is the case of
Ford circles (see [Ford 38]) which arise from integer spinors (see [Ford 38])<fnref n="9" />.

The important point for us, however, is that, in addition to this approach, many other
works such as those of Kocik (see [Kocik 07], [Kocik 19], [Kocik 20¹], [Kocik 20²],
[Kocik 21]) also use spinors to describe Cartesian circle configurations and Apollonian
stackings. In these works, spinors are complex numbers (defined up to the sign)
describing the tangents between pairs of circles. We will show later how Descartes's
theorem is deduced from these spinor considerations.

<footnote n="8">⁸Informally (more details will be provided later), we will say that an *n*-flower is an ordered configuration consisting of a central circle and *n* petal circles all around, such that these are externally tangent to the central circle, and each is externally tangent to the one before and after it.</footnote>

<footnote n="9">⁹"Ford circles" are named in honor of the American mathematician Lester Ford (senior), who described them in a paper published in 1938. A circle is placed on each irreducible fraction, for example 0/1, 1/1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5. Each circle will be tangent to the x-axis as well as to neighboring circles. Fractions with the same denominator have circles of the same size.</footnote>
<figure><img src="image_5.png" /><figcaption>Figure 4: Apollonian Gasket</figcaption></figure>

But we can already draw a provisional conclusion. Against the growing skepticism regarding knowledge, which is indeed changeable because it is obviously situated in history, Bachelard once insisted that certain truths are "forever" in science. And not content with remaining what they are, that is, fixed once and for all, they give rise to generalizations and extensions. For example, Archimedes' principle, which initially applies to fluids, was later applied to gases. Or again, Einstein's theory of relativity, first stated in a restricted form and applied to Galilean motions (rectilinear and uniform), was later generalized to accelerated motions, and then further extended by Weyl and other physicists (see [Parrochia 14]). In other words, there is indeed an "inductive value" of true statements that gradually creates a sanctioned and therefore cumulative history, beyond all paradigm shifts and all imaginable "ruptures."

# 5 Descartes and the Spinors

Let us return, however, to Descartes and the circle theorem. Despite new proofs (including Euler's (see [Euler 90])) (trigonometric) of formula (4), rediscovered, as we have seen, by Steiner, Beecroft, Coxeter, and Pedoe (who listed other proofs), it was not until the early 2000s that the depth of this theorem and its extension possibilities were truly appreciated.
## 5.1 New cases, new writings

From this perspective, the work of Lagarias and his colleagues (see [Lagarias et al. 02] then [Graham et al. 03]) was decisive, paving the way for the later thinking of Kocik (see [Kocik 07]) and then Mathews (see [Mathews-Zymaris 25] and [Mathews 25]) on spinors. Let us briefly review how this generalization was achieved.

First, the introduction of the concept of "curvature" invites us to see that Descartes' work in fact only concerned one of the possible configurations likely to be associated with the circle theorem (see Fig.5).

<figure><img src="image_6.png" /><figcaption>Figure 5: The different configurations of tangent circles according to [Mathews 25]</figcaption></figure>

Descartes only had case (a) in mind, while another circular solution is (b), cases (c) and (d) relating to degenerate tangent circles reduced to straight lines<fnref n="10" />.

$$b_4 + b'_4 = 2(b_1 + b_2 + b_3). \quad (*)$$

Thus, from a Descartes configuration, one can select any of the four circles and replace it with the

<footnote n="10">¹⁰Lagarias (see [Lagarias et al. 02]) observes that given three mutually tangent circles, with curvatures $b_1, b_2$ and $b_3$, there are exactly two other circles tangent to each of them, each giving a four-circle Descartes configuration. The curvatures of these two new circles are the roots of the quadratic equation</footnote>
But the first step in the true generalization is to rewrite the "Cartesian" formula (4) in the form:

$$ \sum_{j=1}^{4} b_j^2 = \frac{1}{2} \left( \sum_{j=1}^{4} b_j \right)^2. $$

then to extend it, via oriented configurations, to spheres of $n$ dimensions, so that we have, at that point, the relation:

$$ \sum_{j=1}^{n+2} b_j^2 = \frac{1}{n} \left( \sum_{j=1}^{n+2} b_j \right)^2. $$

Reiterating the Cartesian construction in each of the circles surrounding the central circle leads to a fractal-like Apollonian stack, sometimes called "Apollonian gasket."¹¹, moreover known to Leibniz (see [Leibniz 76], [Leibniz 06])¹².

But the generalization continues with the introduction of complexes¹³, which allows us to posit, for four mutually tangent circles, of curvature $b_j$ and center $z_j = x_j + iy_j$

$$ \sum_{j=1}^{4} (b_j z_j)^2 = \frac{1}{2} \left( \sum_{j=1}^{4} b_j z_j \right)^2. $$

Lagarias and his colleagues (see [Lagarias et al. 02], 6) were the first to express the Cartesian formula in a matrix form, which Kocik would later reuse. If we write

other circle tangent to the other three; this gives a new Descartes configuration. The new curvature can be obtained from the initial four using (*). This construction can be repeated indefinitely.

¹¹Translated into French as "baderne" by François Apéry (see [Apéry 82]). Starting from the Cartesian configuration, by continuing this construction step by step in this manner, we can add $2 \times 3^n$ new circles at step $n$, giving a total of $3^{n+1} + 2$ circles after $n$ steps. In 1943, it was possible to prove that the residual set, complementary to the disks in the large disk, has a zero area (see [Kasner et al. 43]), and therefore that the stacking is complete. In 1973, it was shown to have a Hausdorff dimension equal to approximately 1.3057.

¹²It is rather curious to note that Michel Serres (see [Serres 68], I, 371), in his famous thesis on the Leibniz system and its mathematical models, does not faithfully reproduce the figure drawn by Leibniz, precisely forgetting the central circle tangent to the other three, and which clearly showed that the diagram was closer to Apollonius and Descartes than to Leeuwenhoeck's microscope and the question of the nesting of the germs with which he seeks to bring it closer. On the other hand, B. Mandelbrot (see [Mandelbrot 82], 243) cites Leibniz's letter to the R.F. des Bosses dated March 11, 1716, as one of the first manifestations of fractal thinking.

¹³In this case, the relation (*) previously mentioned in the footnote transforms into the relation:

$$ b_4 z_4 + b'_4 z'_4 = 2(b_1 z_1 + b_2 z_2 + b_3 z_3). $$

and the general formula including the complex equations takes the form indicated in the text.
Descartes' quadratic form in the generic form:

$$b_1^2 + b_2^2 + b_3^2 + b_4^2 = \frac{1}{2}(b_1 + b_2 + b_3 + b_4)^2,$$

we can make it correspond to the following matrix notation:

$$Q_2 := I_4 - \frac{1}{2} \mathbf{I}_4 \mathbf{I}_4^T = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 & 1 \\ 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \end{bmatrix}$$

where $\mathbf{I}_n$ denotes a column of $n$ 1 and $Q_2$ is the coefficient matrix of the Cartesian quadratic form:

$$Q_2(b_1, b_2, b_3, b_4) := \mathbf{x}^T Q_2 \mathbf{b} = (b_1^2 + b_2^2 + b_3^2 + b_4^2) - \frac{1}{2}(b_1 + b_2 + b_3 + b_4)^2.$$

The subscript 2 in $Q_2$ refers to the dimension of the space considered here. If $\mathbf{b} = (b_1 b_2 b_3 b_4)^T$ denotes the column vector of curvatures and $\mathbf{c} = (b_1 z_1, b_2 z_2, b_3 z_3, b_4 z_4)^T$, then Descartes' theorem simply states that:

$$\mathbf{b}^T Q_2 \mathbf{b} = 0,$$

and Descartes' "complex" theorem that:

$$\mathbf{c}^T Q_2 \mathbf{c} = 0.$$

We can then easily add circles by acting on the matrix, changing the dimension of the quadratic form to express configurations in an $n$-dimensional Euclidean space, and we will then have:

$$Q_n(\mathbf{b}) := \mathbf{b}^T Q_n \mathbf{b} = 0,$$

or move from circles to spheres, and even leave Euclidean space for spherical or hyperbolic geometries where we can also study Apollonian stacks, if desired.

Descartes' theorem is therefore generalized today in all possible ways.<fnref n="14" />

<footnote n="14">¹⁴Let us note in passing that Clifford algebras would allow a much simpler and more elegant proof of Descartes' theorem than the one we gave above. As we have already said, this theorem indeed links the square of a linear form $(\sum k_i)^2$ to a quadratic form $(\sum k_j^2)$. By interpreting the curvatures $k_i$ as the elements of a Clifford algebra, we can encode the relation directly, via anticommutators. Let us only sketch the proof here: let us consider a Clifford algebra $\mathcal{A}$ with generators $e_1, e_2, e_3, e_4$</footnote>
## 5.2 The Transition to Spinors

As we suggested above, as soon as we have an equivalence between a quadratic form and the square of a linear form, we are in fact in the context that led, in the 19th century, to the emergence of Clifford algebras and, in the 20th century, with Dirac, to the expression of the relativistic Hamiltonian of the wave function, the Pauli matrices associated with particle spin, and finally, in their wake, the notion of "spinor," which is introduced quite naturally. We therefore necessarily find the same formalism here, hidden beneath Descartes' theorem.

This time, it was Kocik, initially working on Pythagorean triples (see [Kocik 07] and also [Parrochia 20]), who introduced this particularly illuminating concept, that of the "tangency spinor."

Recall that spinors were introduced by Élie Cartan (1869-1951) in 1913 (see [Cartan 13]) and subsequently named by Paul Ehrenfest (1880-1933) for physical reasons. They have indeed been widely used by quantum mechanics in order to parameterize a new degree of freedom: this one appeared with the terms realizing the equivalence of quadratic forms and squares of linear forms in the Dirac equation, terms associated with matrices expressing a sort of rotation of the particles on themselves, the spin<fnref n="15" />.

Present in Cartan's work as part of a study of linear representations of simple groups, spinors also provide, from a mathematical point of view, a linear representation of the group of rotations of a space with any number of dimensions $n$. Let us briefly introduce them here in the case of two-component spinors:

verifying $e_i^2 = 2$ and $e_i e_j + e_j e_i = 0$ for $i \neq j$. Let us define $K = k_1 e_1 + k_2 e_2 + k_3 e_3 + k_4 e_4$. Let's calculate $K^2$:

$$K^2 = (k_1^2 + k_2^2 + k_3^2 + k_4^2) \cdot 2 + 2(k_1 k_2 e_1 e_2 + \dots).$$

If $K$ is isotropic (i.e $K^2 = 0$), then:

$$\sum k_i^2 = 0 \quad \text{et} \quad \sum k_i k_j e_i e_j = 0.$$

By imposing tangency (relations $k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{\sum k_i k_j}$), we find the theme.

<footnote n="15">¹⁵Each spinor has $2n$ components, depending on whether $n = 2n+1$ or $n = 2n$. When $n = 2$, the spinors of 4-dimensional space appear in the famous Dirac electron equations, the 4 wave functions being in fact the components of a spinor. The wave function of a fermion is therefore represented by a spinor. For particles with spin 1/2 (notably the electron), this is expressed by the Dirac equation. For hypothetical particles with spin 3/2, the Rarita-Schwinger equation would apply. Books abound on the concept of spinors, particularly by physicists and mathematicians. These include [Cartan 38], [Streater et al. 64], [Penrose 84], and [Hladik 96]. But the origin of this concept is geometric.</footnote>
Let two vectors $\mathbf{X}_1 = (x_1, y_1, z_1)$ and $\mathbf{X}_2 = (x_2, y_2, z_2)$ be in the Euclidean space $\mathbb{E}_3$, with the same origin $O$, mutually orthogonal, and of equal norm. These vectors define a plane, and if we consider them as ordered, this order defines a direction of rotation, in this case, that corresponding to the indices. Since the vectors are orthogonal and have the same magnitude, we can write:

$$ \begin{gathered} ||\mathbf{X}_1||^2 = ||\mathbf{X}_2||^2 = x_1^2 + y_1^2 + z_1^2 = x_2^2 + y_2^2 + z_2^2 \\ \mathbf{X}_1 \cdot \mathbf{X}_2 = x_1x_2 + y_1y_2 + z_1z_2 = 0. \end{gathered} \quad (17) $$

To introduce an algebraic representation of the order, we multiply the components of the second vector by the imaginary number $i$, thus forming the three complex numbers:

$$ x = x_1 + ix_2, \quad y = y_1 + iy_2, \quad z = z_1 + iz_2, $$

thus constituting the components of a vector $\mathbf{Z} = (x, y, z)$ which, due to the choice of the previous vectors $\mathbf{X}_1$ and $\mathbf{X}_2$, are not independent, so we can write:

$$ \mathbf{Z} \cdot \mathbf{Z} = ||\mathbf{Z}||^2 = x^2 + y^2 + z^2 = 0, $$

which makes $\mathbf{Z}$, a vector with zero norm, a so-called "isotropic" vector, orthogonal to itself. We can then observe that the relationship between the three complex numbers $x, y$ and $z$ allows them to be expressed using only two complex numbers. Indeed, this relationship can be expressed in the form:

$$ z^2 = -(x^2 + y^2) = -(x + iy)(x - iy), $$

and if we set:

$$ x + iy = -2\phi^2, \quad x - iy = 2\psi^2, $$

these two numbers $\phi$ and $\psi$ allow us to calculate $x, y$ and $z$, since we have:

$$ x = \psi^2 - \phi^2, \quad y = i(\psi^2 - \phi^2), \quad z = \pm 2\psi\phi \quad (\text{we generally choose the - sign}) \quad (18) $$

The set of two complex numbers $\psi$ and $\phi$ constitutes a representation of the two vectors $\mathbf{X}_1$ and $\mathbf{X}_2$ as well as of the order chosen for them. The pair of numbers $(\psi, \phi)$, linked to the vectors $\mathbf{X}_1$ and $\mathbf{X}_2$ by relations (15), constitutes a *spinor*.

We introduced spinors via the 3-dimensional space, but they can exist in any dimension. Moreover, they are linked to the operation equating a quadratic form with a linear form because the coefficients that allow this equivalence, taken from a matrix
algebra called the Clifford algebra, are generated by unitary matrices; any rotation in the reference space causes the spinors, via its associated matrix, to undergo a specific unitary transformation. Note that spinors are stable under addition and multiplication by a complex number, so they constitute intrinsic mathematical entities (expressing in particular the fact that a rotation of $2\pi$ does not return to zero) and form a particular vector space. The same situations being linked to the same formalisms, in the case of Cartesian circles, any ordered pair of disks in the Euclidean plane, which are also capable of undergoing rotations, gives rise to a spinor $u \in \mathbb{R}^2$, an element of the spinor space of dimension 2 (symplectic space), defined up to sign. Algebraically, the process can be described as follows: the disks are represented by unit vectors of space type in the Minkowski space $M = \mathbb{R}^{3,1}$. A certain product of these vectors lies in the null cone of a subspace isomorphic to the Minkowski space $\mathbb{R}^{2,1} \subset M$. Such vectors can in turn be represented by the tensor square of a spinor, an element of the spinor space associated with $\mathbb{R}^{2,1}$.

It is, more specifically, the notion of a "tangency spinor" that will be useful here. A tangency spinor of an ordered pair of mutually tangent disks in the Euclidean plane (conveniently identified with the complex plane $\mathbb{R}^2 \cong \mathbb{C}$) is a vector (concretely, a complex number) such that:

$$u = \pm \sqrt{\frac{z}{r_1 r_2}},$$

where $z = \overrightarrow{O_1O_2}$ is the vector (expressed by the complex number) joining the centers $O_1$ and $O_2$ of the disks of radii $r_1$ and $r_2$, respectively. The tangency spinor is defined up to a sign.

This definition might seem artificial or arbitrary at first glance. However, as Kocik shows, it leads to a number of surprising and fruitful properties. For example, knowing that a quartet of mutually tangent disks is called a "Descartes configuration," Kocik demonstrates that the spinors in a Descartes configuration admit a choice of signs such that the following two properties hold:

$$\text{"rot } \mathbf{u} = 0\text{"} : u_{12} + u_{23} + u_{31} = 0, \qquad (19)$$

$$\text{"div } \mathbf{u} = 0\text{"} : u_{14} + u_{24} + u_{34} = 0 \qquad (20)$$

where $u_{ij}$ represents a spinor for the $i$-th and $j$-th tangent disks as shown in Fig. 6.

Kocik points out that the arrow representation of spinors in the figures is purely symbolic and only indicates the order of the disks. However, the last result mentioned
<figure><img src="image_7.png" /><figcaption>Figure 6: Spinor theorems according to [Kocik 19]</figcaption></figure>

above, "div $\mathbf{u}$ = 0", can be considered a "spinorial" version of Descartes's theorem on circles, from which this one can also be deduced. Let's briefly demonstrate it here.

**Théorème 5.1.** *The relation (20) implies Descartes’ theorem.*

*Proof.* We start with the well-known relation (see, for example, [Deheuvels 81]):

$$||\mathbf{u}\mathbf{v}||^2 + ||\mathbf{u} \times \mathbf{v}||^2 = ||\mathbf{u}||^2||\mathbf{v}||^2,$$

which is in fact, in disguised form, the Pythagorean theorem $cos^2\theta + sin^2\theta = 1$, known in three-dimensional vector calculus and valid in dimension 2. Consider the configuration and notation of Fig. 7.

We have:

$$a + b = d \Rightarrow ||d||^2 = ||a||^2 + 2a \cdot b + |b|^2$$

This gives the meaning of the scalar product:

$$\begin{aligned} 2a \cdot b &= ||d||^2 - ||a||^2 - ||b||^2 \\ &= C + D - (A + C) - B + C \\ &= D - A - B - C. \end{aligned} \qquad (21)$$

On the other hand, we can calculate the scalar product using:

$$|a \cdot b|^2 = |a|^2 |b|^2 - (a \times b)^2 = (A + C)(B + C) - C^2 = AB + BC + CA.$$
<figure><img src="image_8.png" /><figcaption>Figure 7: Proof of proposition 15 according to [Kocik 19]</figcaption></figure>

By equating the scalar product of two equations, we obtain:

$$4(AB + BC + CA)^2 = (D - A - B - C)^2,$$

which, after expansion and grouping of terms, is none other than Descartes’ formula.

<figure><img src="image_9.png" /></figure>

In the case of Apollonian disk stacks, spinors are defined at each point of tangency of two disks, but we will not pursue this technical issue further and will stop our commentary on Kocik here.<fnref n="16" />.

## 5.3 Mathews’s Approach

To the knowledge of Mathews et al., none of the previously cited works provides a generalization of Descartes’ circle theorem to what appear, in his vocabulary, as "n-flowers" for $n > 3$, although there are results involving flower-shaped configurations, such as the Soddy hexlet (see [Soddy 37]). In fact, according to the authors, these works do not provide an explicit equation relating the curvatures.

<footnote n="16">¹⁶The initial motivation for tangency spinors, from his perspective, was the discovery that the Apollonian Gasket (and other integral Apollonian disk stacks) contain Pythagorean triples. Spinors were thus introduced as the geometric representation of their Euclidean parameterizations. The name "spinor" is justified by the fact that these can be considered as spinors of Minkowski space $\mathbb{R}^{2+1}$, in a manner quite analogous to physical spacetime $\mathbb{R}^{3+1}$.</footnote>
Since "flowers" are simple examples of stacks of circles, general stacking theory applies to them (see [Steiner 26]). In general, as Mathews and al. explain, "from a simplicial 2-complex $K$ triangulating an oriented surface, the theory of circle stacking studies the existence and uniqueness of circle stacks realizing $K$, in the sense that the vertices of $K$ correspond to circles, the edges to tangencies and the triangles to triplets of oriented tangent circles. Flowers appear in the simple case where $K$ is a disk constructed from $n$ triangles around a central vertex" (see [Mathews-Zymaris 25], 3). When $K$ is topologically a disk, Stephenson's "boundary value theorem" (see [Steiner 26], thm. 11.6) applies, and the curvatures of the circles at the boundary vertices and the branching structure can be specified arbitrarily. There then exists a unique stacking (in Euclidean or hyperbolic geometry). The work of Mathews et al. then gives the Euclidean curvature of the inner circle in the case of the flower, as a root of an algebraic equation.

Starting from Descartes' classical circle theorem, which, in its modern interpretation, connects the curvatures of four mutually externally tangent circles, three "petal" circles around a central circle, forming a three-flower configuration, Mathews and Zymaris therefore attempt to generalize this theorem to the case of an "n-flower," consisting of $n$ tangent circles around a central circle, in order to give an explicit equation satisfied by their curvatures.

Without going into the details of such a construction (which assumes a spinor description of horospheres in hyperbolic geometry), let us nevertheless give a brief overview, if only to show how inspiring Descartes' theorem remains for contemporary mathematicians.

Let $n \ge 3$. An $n$-flower consists of a central circle $C_\infty$ and $n$ petal circles $C_j$, over integers $j$ mod $n$, such that the $js$ are externally tangent to $C_\infty$ in order around $C_\infty$, and each $C_j$ is externally tangent to $C_{j-1}$ and $C_{j+1}$.

In this article, the authors denote the curvature of a circle $C_\bullet$ by $\kappa_\bullet$. Descartes' circle theorem gives an equation satisfied by the curvatures of a three-dimensional flower, and in this case we have:

$$ (\kappa_{\infty} + \kappa_{1} + \kappa_{2} + \kappa_{3})^{2} = 2(\kappa_{\infty}^{2} + \kappa_{1}^{2} + \kappa_{2}^{3} + \kappa_{3}^{2}). \qquad (22) $$

This therefore requires presenting an explicit equation satisfied by the curvatures of an $n$-dimensional flower. This generalization of Descartes' theorem then takes the following form:

Let $C_\infty$ and $C_j (j \in \mathbb{Z}/n\mathbb{Z})$ be the circles of an $n$-flower having curvatures $\kappa_\infty, \kappa_j$
respectively. The authors define $m_0$ and $m_j$ for $1 \le j \le n-1$ as:

$$m_0 = \sqrt{\frac{K_0}{k_\infty}} + 1, m_j = \sqrt{\left(\frac{K_j}{k_\infty} + 1\right)\left(\frac{K_{j-1}}{k_\infty} + 1\right)} - 1. \quad (23)$$

So, for $n$ odd, we have:

$$\frac{m_0^2}{2} \left( \prod_{j=1}^{n-1} (m_j - i) - \prod_{j=1}^{n-1} (m_j + i) \right) - \prod_{j=1}^{\frac{n-1}{2}} (m_{2j}^2 + 1) = 0. \quad (24)$$

And for $n$ even, the following equation holds:

$$\frac{i}{2} \left( \prod_{j=1}^{n-1} (m_j - i) - \prod_{j=1}^{n-1} (m_j + i) \right) - \prod_{j=1}^{\frac{n-1}{2}} (m_{2j-1}^2 + 1) = 0. \quad (25)$$

In other words, from the curvatures $\kappa_\bullet$, the authors define auxiliary variables $m_\bullet$ via (23), and the $m_\bullet$ satisfy polynomial equations. Since every $\kappa_\bullet > 0$, each $m_\bullet$ is the square root of a clearly positive number, and they take $m_\bullet$ as the positive square root. In (24) and (25), $i$ is the usual square root of $-1$. Although complex numbers are essential to the proof, for writing these equations, they are mostly a convenience. After expansion and cancellation of terms, the resulting polynomials have integer coefficients. Indeed, by writing $[n]$ for $\{1, 2, ..., n\}$, then for any subset $K \subseteq [n-1]$, the term of the product $\prod_{j=1}^{n-1} (m_j \pm i)$ including precisely the $m_k$ with $k \in K$ is given by $(\pm i)^{n-1-|K|}$, $\prod_{k \in K} m_k$. These terms come in pairs, one from $\prod_{j=1}^{n-1} (m_j - i)$ and the other from $\prod_{j=1}^{n-1} (m_j + i)$. When $n-1-|K|$ is even, the terms of each pair are real, equal and cancel each other out; when $n-1-|K|$ is odd, that is, equal to $2l+1$, then the terms of each pair are imaginary and conjugate. Thus, since $\frac{i}{2}((-i)^{n-1-|K|} - i^{n-1-|K|}) = (-1)^l$ (24) and (25) can be written in the form;

$$m_0^2 = \sum_{\substack{K \subseteq [n-1] \\ K \subseteq [n-1] | K| = n-2l-2}} (-1)^l \prod_{k \in K} m_k = \prod_{j=1}^{\frac{n-1}{2}} (m_{2-1j}^2 + 1) \quad (26)$$

And:

$$\sum_{\substack{K \subseteq [n-1] \\ K \subseteq [n-1] | K| = n-2l-2}} (-1)^l \prod_{k \in K} m_k = \prod_{j=1}^{\frac{n-1}{2}} (m_{2j}^2 + 1) \quad (27)$$
respectively.

In any case, making the substitutions of (23) into (24) or (25) (or (26) or (27)) gives an equation satisfied by the $\kappa_{\bullet}$, providing a generalization of Descartes' equation (25). This equation involves square roots, but by multiplying by various conjugates (replacing various $m_{\bullet}$ with $-m_{\bullet}$), we can obtain a polynomial relationship between the $m_{\bullet}^2$; after substitution and erasure of the denominators, we obtain a polynomial relationship between the $\kappa_{\bullet}$. In this way, from equation (25), with $n = 3$, we can find the Cartesian circle theorem. Similarly, from (25), with $n = 4$, we obtain the following equation for the curvatures in a 4-flower:

$$16\kappa_{\infty}^4 - 8\kappa_{\infty}^2(\kappa_1\kappa_2 + \kappa_2\kappa_3 + \kappa_3\kappa_4 + \kappa_4\kappa_1 + 2\kappa_1\kappa_3 + 2\kappa_2\kappa_4) + (\kappa_1^2 + \kappa_3^2)(\kappa_2^2 + \kappa_4^2) \\ - 16\kappa_{\infty}(\kappa_1\kappa_2\kappa_3 + \kappa_2\kappa_3\kappa_4 + \kappa_3\kappa_4\kappa_1 + \kappa_4\kappa_1\kappa_2) - 12\kappa_1\kappa_2\kappa_3\kappa_4 \\ - 2(\kappa_1\kappa_2 + \kappa_3\kappa_4)(\kappa_2\kappa_3 + \kappa_4\kappa_1) = 0. \qquad (28)$$

The larger $n$ is, obviously, the faster the polynomials grow in size and degree.

"Flowers" are thus a building block of circular packing theory. The general theory of circular packing shows that once the curvature of the petals of a flower of type $n$ is known, the curvature of the central circle is determined. Descartes' generalized theorem provides an equation for determining this central curvature.

Based on the work of Mathews (see [Mathews 25], Penrose-Rindler (see [Penrose 84]) and Penner (see [Penner 87]), the demonstration of this result is not obvious and requires some rather subtle reasoning. But its outcome is an incredible generalization of Descartes' theorem, in other words an extension of the simple consideration of 3-flowers to that of $n$-flowers ($n$ tangent circles around the central circle) with an explicit equation satisfied by the curvatures. A final bouquet.

# 6 Conclusion

It may seem futile to the philosopher to be interested in a theorem of Descartes that does not seem to present much interest, a priori, for his philosophy, nor for philosophy in general. No doubt the philosopher exiled in Egmond aan den Hoef (a village near the sea, in the Dutch province of North Holland) was initially interested in it as a mathematician and submitted it to Princess Elizabeth as an exercise. But does one do this kind of thing absolutely gratuitously when one is a philosopher? And does
one reflect on contact without ulterior motive when one admires (if not a little in love with) a person of a completely different background than one's own? How can one find the right distance to be and remain in contact without transgressing the social boundaries of decorum? Throughout this correspondence, the obsequious Descartes lost himself in idle circumlocutions ("the very humble and very obedient servant"...) while his correspondent allowed herself to sign "your very affectionate friend at your service" and, speaking of a man who had spent most of his life pursuing designs of love and ambition, admitted to never having "stronger and more constant" ones with the philosopher than that of this affectionate friendship. Mathematics is generally more direct, and in the eyes of a psychoanalyst, "kissing circles" could well pass, not as always overly hyperbolic metaphors, but, to borrow Viète's phrase, a way of "finely touching" Elizabeth.

But let us put aside these speculations, which are unconfirmed. More seriously, the philosophical question at the center of the relationship between Descartes and Elizabeth, as we know, is that of the union of soul and body. How can there be a union of soul and body if both are substances between which there can only be a difference, by definition, "substantial," that is, real and absolute? But of course, there is a third, the soul-body compound, and, moreover, a point of contact between soul and body, which is, moreover, a little more than a point, a sort of small roundness: the pineal gland. The soul, the body, and the union of the two are therefore like circles manifesting points of tangency between them and also with this famous central pineal gland. Perhaps it was to bring this knowledge of contacts into focus that the philosopher submitted this Apollonian problem to the Princess Palatine? It seems to me less false to consider this than to assume, as has been done, that the tree of philosophy is not a tree, but rather bends back on itself to the point of transplanting the branch of morality into the roots, a bushy or rhizomatic solution that no longer has much to do with Descartes (see [Vaquero 09]). On the other hand, any tree in the mathematical sense can also be considered isomorphic to a series of circles (or ellipses) (see [Parrochia-Neuville 13]): a central circle for the trunk (physics and its metaphysical roots), and osculating circles to represent the different branches (in this case, medicine, mechanics, and morality). This is also a possible interpretation of this Cartesian reflection on the circle theorem.

But beyond Descartes and the history of philosophy, as has been said, the circle theorem has an undeniable epistemological interest. Through his successive generalizations, he testifies to the inescapable fact that certain scientific truths are not only inscribed forever in history, but that they are capable of multiple extensions, and that we must therefore grant them an inductive value: not only are they what
they are (so to speak "auto kath hauto", like Platonic ideas), but they are capable of producing offspring, of generating even vaster and more encompassing truths, which do not, however, contravene the core that gave rise to them, but, on the contrary, confirm it.

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