Descartes' theorem

When three of the four circles are congruent, their centers form an equilateral triangle, as do their points of tangency. The two possibilities for a fourth circle tangent to all three are concentric, and equation (2) reduces to[28]

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

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

If one of the three circles is replaced by a straight line tangent to the remaining circles, then its curvature is zero and drops out of equation (1). For instance, if ${\displaystyle k_{3}=0}$, then equation (1) can be factorized as[29]

$${\displaystyle {\begin{aligned}&{\bigl (}{\sqrt {k_{1}}}+{\sqrt {k_{2}}}+{\sqrt {k_{4}}}{\bigr )}{\bigl (}{{\sqrt {k_{2}}}+{\sqrt {k_{4}}}-{\sqrt {k_{1}}}}{\bigr )}\\[3mu]&\quad {}\cdot {\bigl (}{\sqrt {k_{1}}}+{\sqrt {k_{4}}}-{\sqrt {k_{2}}}{\bigr )}{\bigl (}{\sqrt {k_{1}}}+{\sqrt {k_{2}}}-{\sqrt {k_{4}}}{\bigr )}=0,\end{aligned}}}$$

and equation (2) simplifies to[30]

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

Taking the square root of both sides leads to another alternative formulation of this case (with ${\displaystyle k_{1}\geq k_{2}}$),

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

which has been described as "a sort of demented version of the Pythagorean theorem".[23]

If two circles are replaced by lines, the tangency between the two replaced circles becomes a parallelism between their two replacement lines. In this case, with ${\displaystyle k_{2}=k_{3}=0}$, equation (2) is reduced to the trivial

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

This corresponds to the observation that, for all four curves to remain mutually tangent, the other two circles must be congruent.[17][24]

An Apollonian gasket with integer curvatures, generated by four mutually tangent circles with curvatures −10 (the outer circle), 18, 23, and 27

When four tangent circles described by equation (2) all have integer curvatures, the alternative fourth circle described by the second solution to the equation must also have an integer curvature. This is because both solutions differ from an integer by the square root of an integer, and so either solution can only be an integer if this square root, and hence the other solution, is also an integer. Every four integers that satisfy the equation in Descartes' theorem form the curvatures of four tangent circles.[31] Integer quadruples of this type are also closely related to Heronian triangles, triangles with integer sides and area.[32]

Starting with any four mutually tangent circles, and repeatedly replacing one of the four with its alternative solution (Vieta jumping), in all possible ways, leads to a system of infinitely many tangent circles called an Apollonian gasket. When the initial four circles have integer curvatures, so does each replacement, and therefore all of the circles in the gasket have integer curvatures. Any four tangent circles with integer curvatures belong to exactly one such gasket, uniquely described by its root quadruple of the largest four largest circles and four smallest curvatures. This quadruple can be found, starting from any other quadruple from the same gasket, by repeatedly replacing the smallest circle by a larger one that solves the same Descartes equation, until no such reduction is possible.[31]

A root quadruple is said to be primitive if it has no nontrivial common divisor. Every primitive root quadruple can be found from a factorization of a sum of two squares, ${\displaystyle n^{2}+m^{2}=de}$, as the quadruple ${\displaystyle (-n,\,d+n,\,e+n,\,d+e+n-2m)}$. To be primitive, it must satisfy the additional conditions ${\displaystyle \gcd(n,d,e)=1}$, and ${\displaystyle -n\leq 0\leq 2m\leq d\leq e}$. Factorizations of sums of two squares can be obtained using the sum of two squares theorem. Any other integer Apollonian gasket can be formed by multiplying a primitive root quadruple by an arbitrary integer, and any quadruple in one of these gaskets (that is, any integer solution to the Descartes equation) can be formed by reversing the replacement process used to find the root quadruple. For instance, the gasket with root quadruple ${\displaystyle (-10,18,23,27)}$, shown in the figure, is generated in this way from the factorized sum of two squares ${\displaystyle 10^{2}+2^{2}=8\cdot 13}$.[31]

Ford circles in the unit interval

The special cases of one straight line and integer curvatures combine in the Ford circles. These are an infinite family of circles tangent to the ${\displaystyle x}$-axis of the Cartesian coordinate system at its rational points. Each fraction ${\displaystyle p/q}$ (in lowest terms) has a circle tangent to the line at the point ${\displaystyle (p/q,0)}$ with curvature ${\displaystyle 2q^{2}}$. Three of these curvatures, together with the zero curvature of the axis, meet the conditions of Descartes' theorem whenever the denominators of two of the corresponding fractions sum to the denominator of the third. The two Ford circles for fractions ${\displaystyle p/q}$ and ${\displaystyle r/s}$ (both in lowest terms) are tangent when ${\displaystyle |ps-qr|=1}$. When they are tangent, they form a quadruple of tangent circles with the ${\displaystyle x}$-axis and with the circle for their mediant ${\displaystyle (p+r)/(q+s)}$.[33]

The Ford circles belong to a special Apollonian gasket with root quadruple ${\displaystyle (0,0,1,1)}$, bounded between two parallel lines, which may be taken as the ${\displaystyle x}$-axis and the line ${\displaystyle y=1}$. This is the only Apollonian gasket containing a straight line, and not bounded within a negative-curvature circle. The Ford circles are the circles in this gasket that are tangent to the ${\displaystyle x}$-axis.[31]

Coxeter's loxodromic sequence of tangent circles. Each circle is labeled by an integer i, its position in the sequence; it has radius ρⁱ and curvature ρ⁻ⁱ.

When the four radii of the circles in Descartes' theorem are assumed to be in a geometric progression with ratio ${\displaystyle \rho }$, the curvatures are also in the same progression (in reverse). Plugging this ratio into the theorem gives the equation

$${\displaystyle 2(1+\rho ^{2}+\rho ^{4}+\rho ^{6})=(1+\rho +\rho ^{2}+\rho ^{3})^{2},}$$

which has only one real solution greater than one, the ratio

$${\displaystyle \rho =\varphi +{\sqrt {\varphi }}\approx 2.89005\ ,}$$

where ${\displaystyle \varphi }$ is the golden ratio. If the same progression is continued in both directions, each consecutive four numbers describe circles obeying Descartes' theorem. The resulting double-ended geometric progression of circles can be arranged into a single spiral pattern of tangent circles, called Coxeter's loxodromic sequence of tangent circles. It was first described, together with analogous constructions in higher dimensions, by H. S. M. Coxeter in 1968.[34][35]

Descartes' theorem can be expressed as a matrix equation and then generalized to other configurations of four oriented circles by changing the matrix. Let ${\displaystyle \mathbf {k} }$ be a column vector of the four circle curvatures and let ${\displaystyle \mathbf {Q} }$ be a symmetric matrix whose coefficients ${\displaystyle q_{i,j}}$ represent the relative orientation between the ith and jth oriented circles at their intersection point:

$${\displaystyle \mathbf {Q} ={\begin{bmatrix}{\phantom {-}}1&-1&-1&-1\\-1&{\phantom {-}}1&-1&-1\\-1&-1&{\phantom {-}}1&-1\\-1&-1&-1&{\phantom {-}}1\\\end{bmatrix}},\qquad \mathbf {Q} ^{-1}={\frac {1}{4}}{\begin{bmatrix}{\phantom {-}}1&-1&-1&-1\\-1&{\phantom {-}}1&-1&-1\\-1&-1&{\phantom {-}}1&-1\\-1&-1&-1&{\phantom {-}}1\\\end{bmatrix}}.}$$

Then equation (1) can be rewritten as the matrix equation[17][40]

$${\displaystyle \mathbf {k} ^{\mathsf {T}}\mathbf {Q} ^{-1}\mathbf {k} =0.}$$

As a generalization of Descartes' theorem, a modified symmetric matrix ${\displaystyle \mathbf {Q} }$ can represent any desired configuration of four circles by replacing each coefficient with the inclination ${\displaystyle q_{i,j}}$ between two circles, defined as

$${\displaystyle q_{i,j}={\frac {r_{i}^{2}+r_{j}^{2}-d_{i,j}^{2}}{2r_{i}r_{j}}},}$$

where ${\displaystyle r_{i},r_{j}}$ are the respective radii of the circles, and ${\displaystyle d_{i,j}}$ is the Euclidean distance between their centers.[41][42][43] When the circles intersect, ${\displaystyle q_{i,j}=\cos(\theta _{i,j})}$, the cosine of the intersection angle between the circles. The inclination, sometimes called inversive distance, is ${\displaystyle 1}$ when the circles are tangent and oriented the same way at their point of tangency, ${\displaystyle -1}$ when the two circles are tangent and oriented oppositely at the point of tangency, ${\displaystyle 0}$ for orthogonal circles, outside the interval ${\displaystyle [-1,1]}$ for non-intersecting circles, and ${\displaystyle \infty }$ in the limit as one circle degenerates to a point.[40][35]

The equation ${\displaystyle \mathbf {k} ^{\mathsf {T}}\mathbf {Q} ^{-1}\mathbf {k} =0}$ is satisfied for any arbitrary configuration of four circles in the plane, provided ${\displaystyle \mathbf {Q} }$ is the appropriate matrix of pairwise inclinations.[40]

A special case of Descartes' theorem on the sphere has three circles of radius 60° (k = 1/√3, in blue) for which both circles touching all three (in green) have the same radius (30°, k = √3).

Descartes' theorem generalizes to mutually tangent great or small circles in spherical geometry if the curvature of the ${\displaystyle j}$th circle is defined as ${\textstyle k_{j}=\cot \rho _{j},}$ the cotangent of the oriented intrinsic radius ${\displaystyle \rho _{j}.}$ Then:[42][17]

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

Solving for one of the curvatures in terms of the other three,

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

As a matrix equation,

$${\displaystyle \mathbf {k} ^{\mathsf {T}}\mathbf {Q} ^{-1}\mathbf {k} =-1.}$$

The quantity ${\displaystyle 1/k_{j}=\tan \rho _{j}}$ is the "stereographic diameter" of a small circle. This is the Euclidean length of the diameter in the stereographically projected plane when some point on the circle is projected to the origin. For a great circle, such a stereographic projection is a straight line through the origin, so ${\displaystyle k_{j}=0}$.[44]

Four generalized circles through the origin of the Poincaré disk model of the hyperbolic plane: Circle (blue), horocycle (red), hypercycle (purple), and geodesic (green). The boundary of ideal points is represented with a dashed stroke, and the shaded region is outside the plane.

Likewise, the theorem generalizes to mutually tangent circles in hyperbolic geometry if the curvature of the ${\displaystyle j}$th cycle is defined as ${\textstyle k_{j}=\coth \rho _{j},}$ the hyperbolic cotangent of the oriented intrinsic radius ${\displaystyle \rho _{j}.}$ Then:[17][42]

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

Solving for one of the curvatures in terms of the other three,

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

As a matrix equation,

$${\displaystyle \mathbf {k} ^{\mathsf {T}}\mathbf {Q} ^{-1}\mathbf {k} =1.}$$

This formula also holds for mutually tangent configurations in hyperbolic geometry including hypercycles and horocycles, if ${\displaystyle k_{j}}$ is taken to be the reciprocal of the stereographic diameter of the cycle. This is the diameter under stereographic projection (the Poincaré disk model) when one endpoint of the diameter is projected to the origin.[45] Hypercycles do not have a well-defined center or intrinsic radius and horocycles have an ideal point for a center and infinite intrinsic radius, but ${\displaystyle |k_{j}|>1}$ for a hyperbolic circle, ${\displaystyle |k_{j}|=1}$ for a horocycle, ${\displaystyle |k_{j}|<1}$ for a hypercycle, and ${\displaystyle k_{j}=0}$ for a geodesic.[46]

Soddy's hexlet. Any pair of adjacent green spheres together with the two red spheres and the outer gray sphere satisfy the three-dimensional case of Descartes' theorem.

In ${\displaystyle n}$-dimensional Euclidean space, the maximum number of mutually tangent hyperspheres is ${\displaystyle n+2}$. 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 ${\displaystyle k_{i}=0}$ corresponding to a flat hyperplane, generalizing the 2-dimensional version of the theorem.[17][42] 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 ${\displaystyle n}$ dimensions.[17]

In three dimensions, suppose that three mutually tangent spheres are fixed, and a fourth sphere ${\displaystyle S_{1}}$ is given, tangent to the three fixed spheres. The three-dimensional version of Descartes' theorem can be applied to find a sphere ${\displaystyle S_{2}}$ tangent to ${\displaystyle S_{1}}$ and the fixed spheres, then applied again to find a new sphere ${\displaystyle S_{3}}$ tangent to ${\displaystyle S_{2}}$ and the fixed spheres, and so on. The result is a cyclic sequence of six spheres each tangent to its neighbors in the sequence and to the three fixed spheres, a configuration called Soddy's hexlet, after Soddy's discovery and publication of it in the form of another poem in 1936.[15][47]

Higher-dimensional configurations of mutually tangent hyperspheres in spherical or hyperbolic geometry, with curvatures defined as above, satisfy

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

where ${\displaystyle C=2}$ in spherical geometry and ${\displaystyle C=-2}$ in hyperbolic geometry.[42][17]