Ingeometry,Descartes' theorem states that for every four kissing, or mutuallytangentcircles, the radii of the circles satisfy a certainquadratic equation. By solving this equation, one can construct a fourth circle tangent to three given, mutually tangent circles. The theorem is named afterRené Descartes, who stated it in 1643.

Frederick Soddy's 1936 poemThe Kiss Precise summarizes the theorem in terms of thebends (signed inverse radii) of the four circles:

The sum of the squares of all four bends
Is half the square of their sum^[1]

Special cases of the theorem apply when one or two of the circles is replaced by a straight line (with zero bend) or when the bends areintegers orsquare numbers. A version of the theorem usingcomplex numbers allows the centers of the circles, and not just their radii, to be calculated. With an appropriate definition of curvature, the theorem also applies inspherical geometry andhyperbolic geometry. In higher dimensions, an analogous quadratic equation applies to systems of pairwise tangent spheres or hyperspheres.

Geometrical problems involvingtangent circles have been pondered for millennia. In ancient Greece of the third century BC,Apollonius of Perga devoted an entire book to the topic,Ἐπαφαί [Tangencies]. It has been lost, and is known largely through a description of its contents byPappus of Alexandria and through fragmentary references to it inmedieval Islamic mathematics.^[2] However, Greek geometry was largely focused onstraightedge and compass construction. For instance, theproblem of Apollonius, closely related to Descartes' theorem, asks for the construction of a circle tangent to three given circles which need not themselves be tangent.^[3] Instead, Descartes' theorem is formulated using algebraic relations between numbers describing geometric forms. This is characteristic ofanalytic geometry, a field pioneered byRené Descartes andPierre de Fermat in the first half of the 17th century.^[4]

Descartes discussed the tangent circle problem briefly in 1643, in two letters to PrincessElisabeth of the Palatinate.^[5] Descartes initially posed to the princess the problem of Apollonius. After Elisabeth's partial results revealed that solving the full problem analytically would be too tedious, he simplified the problem to the case in which the three given circles are mutually tangent, and in solving this simplified problem 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.^[6][7] Descartes did not provide the reasoning through which he found this relation.^[8]

Japanese mathematics frequently concerned problems involving circles and their tangencies,^[9] and Japanese mathematicianYamaji Nushizumi stated a form of Descartes' circle theorem in 1751. Like Descartes, he expressed it as a polynomial equation on the radii rather than their curvatures.^[10][11] The special case of this theorem for one straight line and three circles was recorded on a Japanesesangaku tablet from 1824.^[12]

Descartes' theorem was rediscovered in 1826 byJakob Steiner,^[13] in 1842 by Philip Beecroft,^[14] and in 1936 byFrederick Soddy. Soddy chose to format his version of the theorem as a poem,The Kiss Precise, and published it inNature. The kissing circles in this problem are sometimes known asSoddy circles. Soddy also extended the theorem to spheres,^[1] and in another poem described the chain of six spheres each tangent to its neighbors and to three given mutually tangent spheres, a configuration now calledSoddy's hexlet.^[15][16]Thorold Gosset and several others extended the theorem and the poem to arbitrary dimensions; Gosset's version was published the following year.^[17][18] The generalization is sometimes called theSoddy–Gosset theorem,^[19] although both the hexlet and the three-dimensional version were known earlier, in sangaku and in the 1886 work of Robert Lachlan.^[12][20][21] A generalization for multiple kissing circles in two dimensions was proved in 2025.^[22]

Multiple proofs of the theorem have been published. Steiner's proof usesPappus chains andViviani's theorem. Proofs by Philip Beecroft and byH. S. M. Coxeter involve four more circles, passing through triples of tangencies of the original three circles; Coxeter also provided a proof usinginversive geometry. Additional proofs involve arguments based on symmetry, calculations inexterior algebra, or algebraic manipulation ofHeron's formula (for which see§ Soddy circles of a triangle).^[23][24] The result also follows from the observation that theCayley–Menger determinant of the four coplanar circle centers is zero.^[25]

Descartes' theorem is most easily stated in terms of the circles'curvatures.^[26] Thesigned curvature (orbend) of a circle is definedas${\displaystyle k=\pm 1/r}$, where${\displaystyle r}$ is its radius. The larger a circle, the smaller is themagnitude 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 isexternally tangent to the other circles. For aninternally tangent circle thatcircumscribes the other circles, the sign is negative. If a straight line is considered adegenerate circle with zero curvature (and thus infinite radius), Descartes' theorem also applies to a line and three circles that are all three mutually tangent (seeGeneralized circle).^[1]

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:

${\displaystyle (k_1+k_2+k_3+k_4{)}^2=2(k_1^2+k_2^2+k_3^2+k_4^2).}$

${\displaystyle (1)}$

If one of the four curvatures is considered to be a variable, and the rest to be constants, this is aquadratic equation.To find the radius of a fourth circle tangent to three given kissing circles, the quadratic equation can be solved as^[13][27]

${\displaystyle k_4=k_1+k_2+k_3\pm 2\sqrt{k_1{k}_2+k_2{k}_3+k_3{k}_1}.}$

${\displaystyle (2)}$

The${\displaystyle \pm }$ symbol indicates that in general this equation hastwo solutions, and any triple of tangent circles has two tangent circles (or degenerate straight lines). Problem-specific criteria may favor one of these two solutions over the other in any given problem.^[23]

The theorem does not apply to systems of circles with more than two circles tangent to each other at the same point. It requires that the points of tangency be distinct.^[8] When more than two circles are tangent at a single point, there can be infinitely many such circles, with arbitrary curvatures; seepencil of circles.^[28]

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 theCartesian coordinates${\displaystyle (x,y)}$ are interpreted as acomplex number${\displaystyle z=x+iy}$. The equation then looks similar to Descartes' theorem and is therefore called thecomplex Descartes theorem. Given four circles with curvatures${\displaystyle k_i}$ and centers${\displaystyle z_i}$for${\displaystyle i\in \{1,2,3,4\}}$, the following equality holds in addition toequation (1):

${\displaystyle (k_1{z}_1+k_2{z}_2+k_3{z}_3+k_4{z}_4{)}^2=2(k_1^2{z}_1^2+k_2^2{z}_2^2+k_3^2{z}_3^2+k_4^2{z}_4^2).}$

${\displaystyle (3)}$

Once${\displaystyle k_4}$ has been found usingequation (2), one may proceed to calculate${\displaystyle z_4}$ by solvingequation (3) as a quadratic equation, leading to a form similar toequation (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 solutionsfor${\displaystyle z_4}$ corresponding to the two solutionsfor${\displaystyle k_4}$. The plus/minus sign in the above formulafor${\displaystyle z_4}$ does not necessarily correspond to the plus/minus sign in the formulafor${\displaystyle k_4}$.^[19][29][30]

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, andequation (2) reduces to^[31]

$${\displaystyle k_4=(3\pm 2\sqrt{3})k_1.}$$

If one of the three circles is replaced bya straight line tangent to the remaining circles, then its curvature is zero and drops out ofequation (1). For instance,if${\displaystyle k_3=0}$, thenequation (1) can be factorizedas^[32]

andequation (2) simplifiesto^[33]

$${\displaystyle k_4=k_1+k_2\pm 2\sqrt{k_1{k}_2}.}$$

Taking thesquare root of both sides leads to another alternative formulation of this case(with${\displaystyle k_1\ge k_2}$),

$${\displaystyle \sqrt{k_4}=\sqrt{k_1}\pm \sqrt{k_2},}$$

which has been described as "a sort of demented version of thePythagorean theorem".^[26]

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 becongruent.^[19][27]

When four tangent circles described byequation (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 tangentcircles.^[34] Integer quadruples of this type are also closely related toHeronian triangles, triangles with integer sides andarea.^[35]

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 anApollonian 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 itsroot 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.^[34]

A root quadruple is said to beprimitive if it has no nontrivialcommon divisor. Every primitive root quadruple can be found from a factorization of a sum of two squares,${\displaystyle n^2+m^2=de}$, as thequadruple${\displaystyle (-n,d+n,e+n,d+e+n-2m)}$. To be primitive, it must satisfy the additionalconditions${\displaystyle gcd(n,d,e)=1}$,and${\displaystyle -n\le 0\le 2m\le d\le e}$. Factorizations of sums of two squares can be obtained using thesum 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 rootquadruple${\displaystyle (-10,18,23,27)}$, shown in the figure, is generated in this way from the factorized sum of two squares${\displaystyle {10}^2+2^2=8\cdot 13}$.^[34]

The special cases of one straight line and integer curvatures combine in theFord circles. These are an infinite family of circles tangent to the${\displaystyle x}$-axis of theCartesian coordinate system at its rational points. Eachfraction${\displaystyle p/q}$ (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 tangentwhen${\displaystyle |ps-qr|=1}$. When they are tangent, they form a quadruple of tangent circles with the${\displaystyle x}$-axis and with the circle for theirmediant${\displaystyle (p+r)/(q+s)}$.^[36]

The Ford circles belong to a special Apollonian gasket with rootquadruple${\displaystyle (0,0,1,1)}$, bounded between two parallel lines, which may be taken as the${\displaystyle x}$-axis and theline${\displaystyle y=1}$. 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.^[34]

When the four radii of the circles in Descartes' theorem are assumed to be in ageometric progression withratio${\displaystyle \rho }$, 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

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

where${\displaystyle \phi }$ is thegolden 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 singlespiral pattern of tangent circles, calledCoxeter's loxodromic sequence of tangent circles. It was first described, together with analogous constructions in higher dimensions, byH. S. M. Coxeter in 1968.^[37][38]

Anytriangle in the plane has three externally tangent circles centered at its vertices. Letting${\displaystyle A,B,C}$ be the three points,${\displaystyle a,b,c}$ be the lengths of the opposite sides, and$s=\frac{1}{2}(a+b+c)$ be thesemiperimeter, these three circles have radii${\displaystyle s-a,s-b,s-c}$. By Descartes' theorem, two more circles, sometimes calledSoddy circles, are tangent to these three circles. They are separated by theincircle, one interior to it and oneexterior.^[39][40][41] Descartes' theorem can be used to show that the inner Soddy circle's curvatureis$(4R+r+2s)/\mathrm{\Delta }$, where${\displaystyle \mathrm{\Delta }}$ is the triangle's area,${\displaystyle R}$ is itscircumradius, and${\displaystyle r}$ is itsinradius. The outer Soddy circle has curvature$(4R+r-2s)/\mathrm{\Delta }$.^[42] The inner curvature is always positive, but the outer curvature can be positive, negative, or zero. Triangles whose outer circle degenerates to a straight line with curvature zero have been called "Soddyian triangles".^[42]

One of the many proofs of Descartes' theorem is based on this connection to triangle geometry and onHeron's formula for the area of a triangle as a function of its side lengths.If three circles are externally tangent, with radii${\displaystyle r_1,r_2,r_3,}$ then their centers${\displaystyle P_1,P_2,P_3}$ form the vertices of a triangle with side lengths${\displaystyle r_1+r_2,}$${\displaystyle r_1+r_3,}$ and${\displaystyle r_2+r_3,}$ and semiperimeter${\displaystyle r_1+r_2+r_3.}$ By Heron's formula, this triangle${\displaystyle \mathrm{△}{P}_1{P}_2{P}_3}$ has area

$${\displaystyle \sqrt{r_1{r}_2{r}_3(r_1+r_2+r_3)}.}$$

Now consider the inner Soddy circle with radius${\displaystyle r_4,}$ centered at point${\displaystyle P_4}$ inside the triangle. Triangle${\displaystyle \mathrm{△}{P}_1{P}_2{P}_3}$ can be broken into three smaller triangles${\displaystyle \mathrm{△}{P}_1{P}_2{P}_4,}$${\displaystyle \mathrm{△}{P}_4{P}_2{P}_3,}$ and${\displaystyle \mathrm{△}{P}_1{P}_4{P}_3,}$ whose areas can be obtained by substituting${\displaystyle r_4}$ for one of the other radii in the area formula above. The area of the first triangle equals the sum of these three areas:

Careful algebraic manipulation shows that this formula is equivalent toequation (1), Descartes' theorem.^[23]

This analysis covers all cases in which four circles are externally tangent; one is always the inner Soddy circle of the other three. The cases in which one of the circles is internally tangent to the other three and forms their outer Soddy circle are similar. Again the four centers${\displaystyle P_1,P_2,P_3,P_4}$ form four triangles, but (letting${\displaystyle P_4}$ be the center of the outer Soddy circle) the triangle sides incident to${\displaystyle P_4}$ have lengths that are differences of radii,${\displaystyle r_4-r_1,}$${\displaystyle r_4-r_1,}$ and${\displaystyle r_4-r_3,}$ rather than sums.${\displaystyle P_4}$ may lie inside or outside the triangle formed by the other three centers; when it is inside, this triangle's area equals the sum of the other three triangle areas, as above. When it is outside, the quadrilateral formed by the four centers can be subdivided by a diagonal into two triangles, in two different ways, giving an equality between the sum of two triangle areas and the sum of the other two triangle areas. In every case, the area equation reduces to Descartes' theorem. This method does not apply directly to the cases in which one of the circles degenerates to a line, but those can be handled as a limiting case of circles.^[23]

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

Thenequation (1) can be rewritten as thematrix equation^[19][43]

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

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 theinclination${\displaystyle q_{i,j}}$ 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 theEuclidean distance between their centers.^[44][45][46] When the circles intersect,${\displaystyle q_{i,j}=\cos ({\theta }_{i,j})}$, the cosine of the intersection angle between the circles. The inclination, sometimes calledinversive 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,${\displaystyle 0}$ fororthogonal circles, outside the interval${\displaystyle [-1,1]}$ for non-intersecting circles, and${\displaystyle \mathrm{\infty }}$ in the limit as one circle degenerates to apoint.^[43][38]

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.^[43]

Descartes' theorem generalizes to mutually tangentgreat or small circles inspherical geometry if the curvature of the${\displaystyle j}$th circle is defined as$k_j=\cot {\rho }_j,$ thegeodesic curvature of the circle relative to the sphere, which equals thecotangent of the orientedintrinsic radius${\displaystyle {\rho }_j.}$ Then:^[19][45]

$${\displaystyle (k_1+k_2+k_3+k_4{)}^2=2(k_1^2+k_2^2+k_3^2+k_4^2)+4.}$$

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 thestereographically 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${\displaystyle k_j=0}$.^[47]

Likewise, the theorem generalizes to mutually tangentcircles inhyperbolic geometry if the curvature of the${\displaystyle j}$th cycle is defined as$k_j=\coth {\rho }_j,$ the geodesic curvature of the circle relative to the hyperbolic plane, thehyperbolic cotangent of the oriented intrinsic radius${\displaystyle {\rho }_j.}$ Then:^[19][45]

$${\displaystyle (k_1+k_2+k_3+k_4{)}^2=2(k_1^2+k_2^2+k_3^2+k_4^2)-4.}$$

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 includinghypercycles and horocycles, if${\displaystyle k_j}$ is the geodesic curvature of the cycle relative to the hyperbolic plane, the reciprocal of the stereographic diameter of the cycle. This is the diameter under stereographic projection (thePoincaré disk model) when one endpoint of the diameter is projected to the origin.^[48] Hypercycles do not have a well-defined center or intrinsic radius and horocycles have anideal point for a center and infinite intrinsic radius, but${\displaystyle |k_j|>1}$ for a hyperbolic circle,${\displaystyle |k_j|=1}$ for a horocycle, for a hypercycle, and${\displaystyle k_j=0}$ for ageodesic.^[49]

In${\displaystyle n}$-dimensionalEuclidean space, the maximum number of mutually tangenthyperspheres is${\displaystyle n+2}$. For example, in 3-dimensional space, five spheres can be mutually tangent. The curvatures of the hyperspheres satisfy

$${\displaystyle (\sum _{i=1}^{n+2}{k}_i{)}^2=n\sum _{i=1}^{n+2}{k}_i^2}$$

with the case${\displaystyle k_i=0}$ corresponding to a flathyperplane, generalizing the 2-dimensional version of the theorem.^[19][45] Although there is no 3-dimensional analogue of the complex numbers, the relationship between the positions of the centers can be re-expressed as amatrix equation, which also generalizes to${\displaystyle n}$ dimensions.^[19]

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 acyclic sequence of six spheres each tangent to its neighbors in the sequence and to the three fixed spheres, a configuration calledSoddy's hexlet, after Soddy's discovery and publication of it in the form of another poem in 1936.^[15][16]

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

$${\displaystyle (\sum _{i=1}^{n+2}{k}_i{)}^2=nC+n\sum _{i=1}^{n+2}{k}_i^2,}$$

where${\displaystyle C=2}$ in spherical geometry and${\displaystyle C=-2}$ in hyperbolic geometry.^[45][19]

• Circle packing in a circle
• Euler's four-square identity
• Malfatti circles

• Maehara, Hiroshi; Martini, Horst (2024), "Chapter 2: Bend Formulas",Circles, Spheres and Spherical Geometry, Birkhäuser, pp. 31–48,doi:10.1007/978-3-031-62776-7_2,ISBN 978-3-031-62775-0

• Euclidean plane geometry
• Theorems about circles
• Analytic geometry
• Circle packing

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