Descartes' theorem

Geometrical problems involving tangent circles have been pondered for millennia. In ancient Greece of the third century BC, Apollonius of Perga devoted an entire book to the topic, De tactionibus [On tangencies]. It has been lost, and is known only through mentions of it in other works.[1]

René Descartes discussed the problem briefly in 1643, in a letter to Princess Elisabeth of the Palatinate. He came up with the equation describing the relation between the radii, or curvatures, of four pairwise tangent circles. This result became known as Descartes' theorem.[2]

This result was rediscovered in 1826 by Jakob Steiner,[3] in 1842 by Philip Beecroft,[4] and in 1936 by Frederick Soddy. The kissing circles in this problem are sometimes known as Soddy circles and line connecting their centers as Soddy line perhaps because Soddy chose to publish his version of the theorem in the form of a poem, titled The Kiss Precise. Soddy also extended the theorem to spheres;[5] Thorold Gosset extended the theorem to arbitrary dimensions.[6]

Kissing circles. Given three mutually tangent circles (black), what radius can a fourth tangent circle have? There are in general two possible answers (red).

Descartes' theorem is most easily stated in terms of the circles' curvatures. The curvature (or bend) of a circle is defined as ${\displaystyle k=\pm 1/r}$, where ${\displaystyle r}$ is its radius. The larger a circle, the smaller is the magnitude of its curvature, and vice versa.

The sign in ${\displaystyle k=\pm 1/r}$ (represented by the ${\displaystyle \pm }$ symbol) is positive for a circle that is externally tangent to the other circles, like the three black circles in the image. For an internally tangent circle like the large red circle, that circumscribes the other circles, the sign is negative. If a straight line is considered a degenerate circle with zero curvature (and thus infinite radius), Descartes' theorem also applies to a line and three circles that are all three mutually tangent.

For four circles that are tangent to each other at six distinct points, with curvatures ${\displaystyle k_{i}}$ for ${\displaystyle i=1,\dots 4}$, Descartes' theorem says:

${\displaystyle (k_{1}+k_{2}+k_{3}+k_{4})^{2}=2\,(k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}).}$

(1)

To find the radius of a fourth circle tangent to three given kissing circles, the equation can be written

${\displaystyle k_{4}=k_{1}+k_{2}+k_{3}\pm 2{\sqrt {k_{1}k_{2}+k_{2}k_{3}+k_{3}k_{1}}}.}$

(2)

The ± symbol indicates that in general there are two solutions to this equation, and two tangent circles (or degenerate straight lines) to any triple of tangent circles. Problem-specific criteria may favor one of these two solutions over the other in any given problem.

One of the circles is replaced by a straight line of zero curvature. Descartes' theorem still applies.

Here, as all three circles are tangent to each other at the same point, Descartes' theorem does not apply.

If one of the three circles, say number 3, is replaced by a straight line, then its curvature k₃ is zero and drops out of equation (1). Equation (2) then simplifies to:

${\displaystyle k_{4}=k_{1}+k_{2}\pm 2{\sqrt {k_{1}k_{2}}}.}$

(3)

If two circles are replaced by lines, the tangency between the two replaced circles becomes a parallelism between their two replacement lines. For all four curves to remain mutually tangent, the other two circles must be congruent. In this case, with k₂ = k₃ = 0, equation (2) is reduced to the trivial

${\displaystyle \displaystyle k_{4}=k_{1}.}$

It is not possible to replace three circles by lines, as it is not possible for three lines and one circle to be mutually tangent. Descartes' theorem does not apply when all four circles are tangent to each other at the same point.

Another special case is when the k_i are squares,

${\displaystyle (v^{2}+x^{2}+y^{2}+z^{2})^{2}=2\,(v^{4}+x^{4}+y^{4}+z^{4})}$

Euler showed that this is equivalent to the simultaneous triplet of Pythagorean triples,

${\displaystyle (2vx)^{2}+(2yz)^{2}=\,(v^{2}+x^{2}-y^{2}-z^{2})^{2}}$

${\displaystyle (2vy)^{2}+(2xz)^{2}=\,(v^{2}-x^{2}+y^{2}-z^{2})^{2}}$

${\displaystyle (2vz)^{2}+(2xy)^{2}=\,(v^{2}-x^{2}-y^{2}+z^{2})^{2}}$

and can be given a parametric solution. When the minus sign of a curvature is chosen,

${\displaystyle (-v^{2}+x^{2}+y^{2}+z^{2})^{2}=2\,(v^{4}+x^{4}+y^{4}+z^{4})}$

this can be solved[7] as,

${\displaystyle {\begin{aligned}{[}&v,x,y,z]\\[6pt]={}{\Big [}&2(ab-cd)(ab+cd),\ (a^{2}+b^{2}+c^{2}+d^{2})(a^{2}-b^{2}+c^{2}-d^{2}),\\&\qquad 2(ac-bd)(a^{2}+c^{2}),\ 2(ac-bd)(b^{2}+d^{2}){\Big ]}\end{aligned}}}$

where

${\displaystyle a^{4}+b^{4}=\,c^{4}+d^{4}}$

parametric solutions of which are well-known.

To determine a circle completely, not only its radius (or curvature), but also its center must be known. The relevant equation is expressed most clearly if the coordinates (x, y) are interpreted as a complex number z = x + iy. The equation then looks similar to Descartes' theorem and is therefore called the complex Descartes theorem.

Given four circles with curvatures k_i and centers z_i (for i = 1, 2, 3, 4), the following equality holds in addition to equation (1):

${\displaystyle (k_{1}z_{1}+k_{2}z_{2}+k_{3}z_{3}+k_{4}z_{4})^{2}=2\,(k_{1}^{2}z_{1}^{2}+k_{2}^{2}z_{2}^{2}+k_{3}^{2}z_{3}^{2}+k_{4}^{2}z_{4}^{2}).}$

(4)

Once k₄ has been found using equation (2), one may proceed to calculate z₄ by rewriting equation (4) to a form similar to equation (2):

${\displaystyle z_{4}={\frac {z_{1}k_{1}+z_{2}k_{2}+z_{3}k_{3}\pm 2{\sqrt {k_{1}k_{2}z_{1}z_{2}+k_{2}k_{3}z_{2}z_{3}+k_{1}k_{3}z_{1}z_{3}}}}{k_{4}}}.}$

Again, in general there are two solutions for z₄ corresponding to the two solutions for k₄. Note that the plus/minus sign in the above formula for z does not necessarily correspond to the plus/minus sign in the formula for k.

The generalization to n dimensions is sometimes referred to as the Soddy–Gosset theorem, although its three-dimensional case was stated by Robert Lachlan in 1886.[8][9] In n-dimensional Euclidean space, the maximum number of mutually tangent (n − 1)-spheres is n + 2. For example, in 3-dimensional space, five spheres can be mutually tangent. The curvatures of the hyperspheres satisfy

${\displaystyle {\biggl (}\sum _{i=1}^{n+2}k_{i}{\biggr )}^{\!2}=n\,\sum _{i=1}^{n+2}k_{i}^{2}}$

with the case k_i = 0 corresponding to a flat hyperplane, in exact analogy to the 2-dimensional version of the theorem.[8]

Although there is no 3-dimensional analogue of the complex numbers, the relationship between the positions of the centers can be re-expressed as a matrix equation, which also generalizes to n dimensions.[8]

• Ford circles
• Apollonian gasket
• Problem of Apollonius ("circle tangencies")
• Soddy's hexlet
• Tangent lines to circles
• Isoperimetric point

• Interactive applet demonstrating four mutually tangent circles at cut-the-knot
• The Kiss Precise