InEuclidean plane geometry,Apollonius's problem is to construct circles that aretangent to three given circles in a plane (Figure 1).Apollonius of Perga (c. 262BC – c. 190 BC) posed and solved this famous problem in his workἘπαφαί (Epaphaí, "Tangencies"); this work has beenlost, but a 4th-century AD report of his results byPappus of Alexandria has survived. Three given circles generically have eight different circles that are tangent to them (Figure 2), a pair of solutions for each way to divide the three given circles in two subsets (there are 4 ways to divide a set ofcardinality 3 in 2 parts).

In the 16th century,Adriaan van Roomen solved the problem using intersectinghyperbolas, but this solution uses methods not limited tostraightedge and compass constructions.François Viète found a straightedge and compass solution by exploitinglimiting cases: any of the three given circles can be shrunk to zero radius (a point) or expanded to infinite radius (a line). Viète's approach, which uses simpler limiting cases to solve more complicated ones, is considered a plausible reconstruction of Apollonius' method. The method of van Roomen was simplified byIsaac Newton, who showed that Apollonius' problem is equivalent to finding a position from the differences of its distances to three known points. This has applications in navigation and positioning systems such asLORAN.

Later mathematicians introduced algebraic methods, which transform a geometric problem intoalgebraic equations. These methods were simplified by exploitingsymmetries inherent in the problem of Apollonius: for instance solution circles generically occur in pairs, with one solution enclosing the given circles that the other excludes (Figure 2).Joseph Diaz Gergonne used this symmetry to provide an elegant straightedge and compass solution, while other mathematicians usedgeometrical transformations such asreflection in a circle to simplify the configuration of the given circles. These developments provide a geometrical setting for algebraic methods (usingLie sphere geometry) and a classification of solutions according to 33 essentially different configurations of the given circles.

Apollonius' problem has stimulated much further work. Generalizations to three dimensions—constructing a sphere tangent to four given spheres—andbeyond have been studied. The configuration of three mutually tangent circles has received particular attention.René Descartes gave a formula relating the radii of the solution circles and the given circles, now known asDescartes' theorem. Solving Apollonius' problem iteratively in this case leads to theApollonian gasket, which is one of the earliestfractals to be described in print, and is important innumber theory viaFord circles and theHardy–Littlewood circle method.

The general statement of Apollonius' problem is to construct one or more circles that are tangent to three given objects in a plane, where an object may be a line, a point or a circle of any size.^[1][2][3][4] These objects may be arranged in any way and may cross one another; however, they are usually taken to be distinct, meaning that they do not coincide. Solutions to Apollonius' problem are sometimes calledApollonius circles, although the term is also used forother types of circles associated with Apollonius.

The property of tangency is defined as follows. First, a point, line or circle is assumed to be tangent to itself; hence, if a given circle is already tangent to the other two given objects, it is counted as a solution to Apollonius' problem. Two distinct geometrical objects are said tointersect if they have a point in common. By definition, a point is tangent to a circle or a line if it intersects them, that is, if it lies on them; thus, two distinct points cannot be tangent. If the angle between lines or circles at an intersection point is zero, they are said to betangent; the intersection point is called atangent point or apoint of tangency. (The word "tangent" derives from theLatinpresent participle,tangens, meaning "touching".) In practice, two distinct circles are tangent if they intersect at only one point; if they intersect at zero or two points, they are not tangent. The same holds true for a line and a circle. Two distinct lines cannot be tangent in the plane, although twoparallel lines can be considered as tangent at apoint at infinity ininversive geometry (seebelow).^[5][6]

The solution circle may be either internally or externally tangent to each of the given circles. Anexternal tangency is one where the two circles bend away from each other at their point of contact; they lie on opposite sides of thetangent line at that point, and they exclude one another. The distance between their centers equals the sum of their radii. By contrast, aninternal tangency is one in which the two circles curve in the same way at their point of contact; the two circles lie on the same side of the tangent line, and one circle encloses the other. In this case, the distance between their centers equals the difference of their radii. As an illustration, in Figure 1, the pink solution circle is internally tangent to the medium-sized given black circle on the right, whereas it is externally tangent to the smallest and largest given circles on the left.

Apollonius' problem can also be formulated as the problem of locating one or more points such that thedifferences of its distances to three given points equal three known values. Consider a solution circle of radiusr_s and three given circles of radiir₁,r₂ andr₃. If the solution circle is externally tangent to all three given circles, the distances between the center of the solution circle and the centers of the given circles equald₁ =r₁ +r_s,d₂ =r₂ +r_s andd₃ =r₃ +r_s, respectively. Therefore, differences in these distances are constants, such asd₁ −d₂ =r₁ −r₂; they depend only on the known radii of the given circles and not on the radiusr_s of the solution circle, which cancels out. This second formulation of Apollonius' problem can be generalized to internally tangent solution circles (for which the center-center distance equals the difference of radii), by changing the corresponding differences of distances to sums of distances, so that the solution-circle radiusr_s again cancels out. The re-formulation in terms of center-center distances is useful in thesolutions below ofAdriaan van Roomen andIsaac Newton, and also inhyperbolic positioning or trilateration, which is the task of locating a position from differences in distances to three known points. For example, navigation systems such asLORAN identify a receiver's position from the differences in arrival times of signals from three fixed positions, which correspond to the differences in distances to those transmitters.^[7][8]

A rich repertoire of geometrical and algebraic methods have been developed to solve Apollonius' problem,^[9][10] which has been called "the most famous of all" geometry problems.^[3] The original approach ofApollonius of Perga has been lost, but reconstructions have been offered byFrançois Viète and others, based on the clues in the description byPappus of Alexandria.^[11][12] The first new solution method was published in 1596 byAdriaan van Roomen, who identified the centers of the solution circles as the intersection points of twohyperbolas.^[13][14] Van Roomen's method was refined in 1687 byIsaac Newton in hisPrincipia,^[15][16] and byJohn Casey in 1881.^[17]

Although successful in solving Apollonius' problem, van Roomen's method has a drawback. A prized property in classicalEuclidean geometry is the ability to solve problems using only acompass and a straightedge.^[18] Many constructions are impossible using only these tools, such asdividing an angle in three equal parts. However, many such "impossible" problems can be solved by intersecting curves such as hyperbolas,ellipses andparabolas (conic sections). For example,doubling the cube (the problem of constructing a cube of twice the volume of a given cube) cannot be done using only a straightedge and compass, butMenaechmus showed that the problem can be solved by using the intersections of twoparabolas.^[19] Therefore, van Roomen's solution—which uses the intersection of two hyperbolas—did not determine if the problem satisfied the straightedge-and-compass property.

Van Roomen's friendFrançois Viète, who had urged van Roomen to work on Apollonius' problem in the first place, developed a method that used only compass and straightedge.^[20] Prior to Viète's solution,Regiomontanus doubted whether Apollonius' problem could be solved by straightedge and compass.^[21] Viète first solved some simple special cases of Apollonius' problem, such as finding a circle that passes through three given points which has only one solution if the points are distinct; he then built up to solving more complicated special cases, in some cases by shrinking or swelling the given circles.^[1] According to the 4th-century report of Pappus, Apollonius' own book on this problem—entitledἘπαφαί (Epaphaí, "Tangencies"; Latin:De tactionibus,De contactibus)—followed a similar progressive approach.^[11] Hence, Viète's solution is considered to be a plausible reconstruction of Apollonius' solution, although other reconstructions have been published independently by three different authors.^[22]

Several other geometrical solutions to Apollonius' problem were developed in the 19th century. The most notable solutions are those ofJean-Victor Poncelet (1811)^[23] and ofJoseph Diaz Gergonne (1814).^[24] Whereas Poncelet's proof relies onhomothetic centers of circles and thepower of a point theorem, Gergonne's method exploits the conjugate relation between lines and theirpoles in a circle. Methods usingcircle inversion were pioneered byJulius Petersen in 1879;^[25] one example is the annular solution method ofHSM Coxeter.^[2] Another approach usesLie sphere geometry,^[26] which was developed bySophus Lie.

Algebraic solutions to Apollonius' problem were pioneered in the 17th century byRené Descartes andPrincess Elisabeth of Bohemia, although their solutions were rather complex.^[9] Practical algebraic methods were developed in the late 18th and 19th centuries by several mathematicians, includingLeonhard Euler,^[27]Nicolas Fuss,^[9]Carl Friedrich Gauss,^[28]Lazare Carnot,^[29] andAugustin Louis Cauchy.^[30]

The solution ofAdriaan van Roomen (1596) is based on the intersection of twohyperbolas.^[13][14] Let the given circles be denoted asC₁,C₂ andC₃. Van Roomen solved the general problem by solving a simpler problem, that of finding the circles that are tangent totwo given circles, such asC₁ andC₂. He noted that the center of a circle tangent to both given circles must lie on ahyperbola whose foci are the centers of the given circles. To understand this, let the radii of the solution circle and the two given circles be denoted asr_s,r₁ andr₂, respectively (Figure 3). The distanced₁ between the centers of the solution circle andC₁ is eitherr_s +r₁ orr_s −r₁, depending on whether these circles are chosen to be externally or internally tangent, respectively. Similarly, the distanced₂ between the centers of the solution circle andC₂ is eitherr_s +r₂ orr_s −r₂, again depending on their chosen tangency. Thus, the differenced₁ −d₂ between these distances is always a constant that is independent ofr_s. This property, of having a fixed difference between the distances to thefoci, characterizes hyperbolas, so the possible centers of the solution circle lie on a hyperbola. A second hyperbola can be drawn for the pair of given circlesC₂ andC₃, where the internal or external tangency of the solution andC₂ should be chosen consistently with that of the first hyperbola. An intersection of these two hyperbolas (if any) gives the center of a solution circle that has the chosen internal and external tangencies to the three given circles. The full set of solutions to Apollonius' problem can be found by considering all possible combinations of internal and external tangency of the solution circle to the three given circles.

Isaac Newton (1687) refined van Roomen's solution, so that the solution-circle centers were located at the intersections of a line with a circle.^[15] Newton formulates Apollonius' problem as a problem intrilateration: to locate a pointZ from three given pointsA,B andC, such that the differences in distances fromZ to the three given points have known values.^[31] These four points correspond to the center of the solution circle (Z) and the centers of the three given circles (A,B andC).

Instead of solving for the two hyperbolas, Newton constructs theirdirectrix lines instead. For any hyperbola, the ratio of distances from a pointZ to a focusA and to the directrix is a fixed constant called theeccentricity. The two directrices intersect at a pointT, and from their two known distance ratios, Newton constructs a line passing throughT on whichZ must lie. However, the ratio of distances TZ/TA is also known; hence,Z also lies on a known circle, since Apollonius had shown that acircle can bedefined as the set of points that have a given ratio of distances to two fixed points. (As an aside, this definition is the basis ofbipolar coordinates.) Thus, the solutions to Apollonius' problem are the intersections of a line with a circle.

As describedbelow, Apollonius' problem has ten special cases, depending on the nature of the three given objects, which may be a circle (C), line (L) or point (P). By custom, these ten cases are distinguished by three letter codes such asCCP.^[32] Viète solved all ten of these cases using only compass and straightedge constructions, and used the solutions of simpler cases to solve the more complex cases.^[1][20]

Viète began by solving thePPP case (three points) following the method ofEuclid in hisElements. From this, he derived alemma corresponding to thepower of a point theorem, which he used to solve theLPP case (a line and two points). Following Euclid a second time, Viète solved theLLL case (three lines) using theangle bisectors. He then derived a lemma for constructing the line perpendicular to an angle bisector that passes through a point, which he used to solve theLLP problem (two lines and a point). This accounts for the first four cases of Apollonius' problem, those that do not involve circles.

To solve the remaining problems, Viète exploited the fact that the given circles and the solution circle may be re-sized in tandem while preserving their tangencies (Figure 4). If the solution-circle radius is changed by an amount Δr, the radius of its internally tangent given circles must be likewise changed by Δr, whereas the radius of its externally tangent given circles must be changed by −Δr. Thus, as the solution circle swells, the internally tangent given circles must swell in tandem, whereas the externally tangent given circles must shrink, to maintain their tangencies.

Viète used this approach to shrink one of the given circles to a point, thus reducing the problem to a simpler, already solved case. He first solved theCLL case (a circle and two lines) by shrinking the circle into a point, rendering it anLLP case. He then solved theCLP case (a circle, a line and a point) using three lemmas. Again shrinking one circle to a point, Viète transformed theCCL case into aCLP case. He then solved theCPP case (a circle and two points) and theCCP case (two circles and a point), the latter case by two lemmas. Finally, Viète solved the generalCCC case (three circles) by shrinking one circle to a point, rendering it aCCP case.

Apollonius' problem can be framed as a system of three equations for the center and radius of the solution circle.^[33] Since the three given circles and any solution circle must lie in the same plane, their positions can be specified in terms of the (x,y)coordinates of their centers. For example, the center positions of the three given circles may be written as (x₁,y₁), (x₂,y₂) and (x₃,y₃), whereas that of a solution circle can be written as (x_s,y_s). Similarly, the radii of the given circles and a solution circle can be written asr₁,r₂,r₃ andr_s, respectively. The requirement that a solution circle must exactly touch each of the three given circles can be expressed as threecoupledquadratic equations forx_s,y_s andr_s:

${\displaystyle {\left(x_s-x_1\right)}^2+{\left(y_s-y_1\right)}^2={\left(r_s-s_1{r}_1\right)}^2}$

${\displaystyle {\left(x_s-x_2\right)}^2+{\left(y_s-y_2\right)}^2={\left(r_s-s_2{r}_2\right)}^2}$

${\displaystyle {\left(x_s-x_3\right)}^2+{\left(y_s-y_3\right)}^2={\left(r_s-s_3{r}_3\right)}^2.}$

The three numberss₁,s₂ ands₃ on theright-hand side, called signs, may equal ±1, and specify whether the desired solution circle should touch the corresponding given circle internally (s = 1) or externally (s = −1). For example, in Figures 1 and 4, the pink solution is internally tangent to the medium-sized given circle on the right and externally tangent to the smallest and largest given circles on the left; if the given circles are ordered by radius, the signs for this solution are"− + −". Since the three signs may be chosen independently, there are eight possible sets of equations(2 × 2 × 2 = 8), each set corresponding to one of the eight types of solution circles.

The general system of three equations may be solved by the method ofresultants. When multiplied out, all three equations havex_s² +y_s² on the left-hand side, andr_s² on the right-hand side. Subtracting one equation from another eliminates these quadratic terms; the remaining linear terms may be re-arranged to yield formulae for the coordinatesx_s andy_s

${\displaystyle x_s=M+N{r}_s}$

${\displaystyle y_s=P+Q{r}_s}$

whereM,N,P andQ are known functions of the given circles and the choice of signs. Substitution of these formulae into one of the initial three equations gives a quadratic equation forr_s, which can be solved by thequadratic formula. Substitution of the numerical value ofr_s into the linear formulae yields the corresponding values ofx_s andy_s.

The signss₁,s₂ ands₃ on the right-hand sides of the equations may be chosen in eight possible ways, and each choice of signs gives up to two solutions, since the equation forr_s isquadratic. This might suggest (incorrectly) that there are up to sixteen solutions of Apollonius' problem. However, due to a symmetry of the equations, if (r_s,x_s,y_s) is a solution, with signss_i, then so is (−r_s,x_s,y_s), with opposite signs −s_i, which represents the same solution circle. Therefore, Apollonius' problem has at most eight independent solutions (Figure 2). One way to avoid this double-counting is to consider only solution circles with non-negative radius.

The two roots of any quadratic equation may be of three possible types: two differentreal numbers, two identical real numbers (i.e., a degenerate double root), or a pair ofcomplex conjugate roots. The first case corresponds to the usual situation; each pair of roots corresponds to a pair of solutions that are related bycircle inversion, as described below (Figure 6). In the second case, both roots are identical, corresponding to a solution circle that transforms into itself under inversion. In this case, one of the given circles is itself a solution to the Apollonius problem, and the number of distinct solutions is reduced by one. The third case of complex conjugate radii does not correspond to a geometrically possible solution for Apollonius' problem, since a solution circle cannot have an imaginary radius; therefore, the number of solutions is reduced by two. Apollonius' problem cannot have seven solutions, although it may have any other number of solutions from zero to eight.^[12][34]

The same algebraic equations can be derived in the context ofLie sphere geometry.^[26] That geometry represents circles, lines and points in a unified way, as a five-dimensional vectorX = (v,c_x,c_y,w,sr), wherec = (c_x,c_y) is the center of the circle, andr is its (non-negative) radius. Ifr is not zero, the signs may be positive or negative; for visualization,s represents theorientation of the circle, with counterclockwise circles having a positives and clockwise circles having a negatives. The parameterw is zero for a straight line, and one otherwise.

In this five-dimensional world, there is abilinear product similar to thedot product:

${\displaystyle \left(X_1\mid X_2\right):=v_1{w}_2+v_2{w}_1+{\mathbf{c}}_1\cdot {\mathbf{c}}_2-s_1{s}_2{r}_1{r}_2.}$

TheLie quadric is defined as those vectors whose product with themselves (theirsquare norm) is zero, (X|X) = 0. LetX₁ andX₂ be two vectors belonging to this quadric; the norm of their difference equals

${\displaystyle \left(X_1-X_2\mid X_1-X_2\right)=2\left(v_1-v_2\right)\left(w_1-w_2\right)+\left({\mathbf{c}}_1-{\mathbf{c}}_2\right)\cdot \left({\mathbf{c}}_1-{\mathbf{c}}_2\right)-{\left(s_1{r}_1-s_2{r}_2\right)}^2.}$

The productdistributes over addition and subtraction (more precisely, it isbilinear):

${\displaystyle \left(X_1-X_2\mid X_1-X_2\right)=\left(X_1\mid X_1\right)-2\left(X_1\mid X_2\right)+\left(X_2\mid X_2\right).}$

Since (X₁|X₁) = (X₂|X₂) = 0 (both belong to the Lie quadric) and sincew₁ =w₂ = 1 for circles, the product of any two such vectors on the quadric equals

${\displaystyle -2\left(X_1\mid X_2\right)={\left|{\mathbf{c}}_1-{\mathbf{c}}_2\right|}^2-{\left(s_1{r}_1-s_2{r}_2\right)}^2.}$

where the vertical bars sandwichingc₁ −c₂ represent the length of that difference vector, i.e., theEuclidean norm. This formula shows that if two quadric vectorsX₁ andX₂ are orthogonal (perpendicular) to one another—that is, if (X₁|X₂) = 0—then their corresponding circles are tangent. For if the two signss₁ ands₂ are the same (i.e. the circles have the same "orientation"), the circles are internally tangent; the distance between their centers equals thedifference in the radii

${\displaystyle {\left|{\mathbf{c}}_1-{\mathbf{c}}_2\right|}^2={\left(r_1-r_2\right)}^2.}$

Conversely, if the two signss₁ ands₂ are different (i.e. the circles have opposite "orientations"), the circles are externally tangent; the distance between their centers equals thesum of the radii

${\displaystyle {\left|{\mathbf{c}}_1-{\mathbf{c}}_2\right|}^2={\left(r_1+r_2\right)}^2.}$

Therefore, Apollonius' problem can be re-stated in Lie geometry as a problem of finding perpendicular vectors on the Lie quadric; specifically, the goal is to identify solution vectorsX_sol that belong to the Lie quadric and are also orthogonal (perpendicular) to the vectorsX₁,X₂ andX₃ corresponding to the given circles.

${\displaystyle \left(X_{\mathrm{s}\mathrm{o}\mathrm{l}}\mid X_{\mathrm{s}\mathrm{o}\mathrm{l}}\right)=\left(X_{\mathrm{s}\mathrm{o}\mathrm{l}}\mid X_1\right)=\left(X_{\mathrm{s}\mathrm{o}\mathrm{l}}\mid X_2\right)=\left(X_{\mathrm{s}\mathrm{o}\mathrm{l}}\mid X_3\right)=0}$

The advantage of this re-statement is that one can exploit theorems fromlinear algebra on the maximum number oflinearly independent, simultaneously perpendicular vectors. This gives another way to calculate the maximum number of solutions and extend the theorem to higher-dimensional spaces.^[26][35]

A natural setting for problem of Apollonius isinversive geometry.^[4][12] The basic strategy of inversive methods is to transform a given Apollonius problem into another Apollonius problem that is simpler to solve; the solutions to the original problem are found from the solutions of the transformed problem by undoing the transformation. Candidate transformations must change one Apollonius problem into another; therefore, they must transform the given points, circles and lines to other points, circles and lines, and no other shapes.Circle inversion has this property and allows the center and radius of the inversion circle to be chosen judiciously. Other candidates include theEuclidean plane isometries; however, they do not simplify the problem, since they merelyshift,rotate, andmirror the original problem.

Inversion in a circle with centerO and radiusR consists of the following operation (Figure 5): every pointP is mapped into a new pointP' such thatO,P, andP' are collinear, and the product of the distances ofP andP' to the centerO equal the radiusR squared

${\displaystyle \overline{\mathbf{O}\mathbf{P}}\cdot \overline{\mathbf{O}{\mathbf{P}}^{\mathrm{\prime }}}=R^2.}$

Thus, ifP lies outside the circle, thenP' lies within, and vice versa. WhenP is the same asO, the inversion is said to sendP to infinity. (Incomplex analysis, "infinity" is defined in terms of theRiemann sphere.) Inversion has the useful property that lines and circles are always transformed into lines and circles, and points are always transformed into points. Circles are generally transformed into other circles under inversion; however, if a circle passes through the center of the inversion circle, it is transformed into a straight line, and vice versa. Importantly, if a circle crosses the circle of inversion at right angles (intersects perpendicularly), it is left unchanged by the inversion; it is transformed into itself.

Circle inversions correspond to a subset ofMöbius transformations on theRiemann sphere. The planar Apollonius problem can be transferred to the sphere by aninverse stereographic projection; hence, solutions of the planar Apollonius problem also pertain to its counterpart on the sphere. Other inversive solutions to the planar problem are possible besides the common ones described below.^[36]

Solutions to Apollonius's problem generally occur in pairs; for each solution circle, there is a conjugate solution circle (Figure 6).^[1] One solution circle excludes the given circles that are enclosed by its conjugate solution, and vice versa. For example, in Figure 6, one solution circle (pink, upper left) encloses two given circles (black), but excludes a third; conversely, its conjugate solution (also pink, lower right) encloses that third given circle, but excludes the other two. The two conjugate solution circles are related byinversion, by the following argument.

In general, any three distinct circles have a unique circle—theradical circle—that intersects all of them perpendicularly; the center of that circle is theradical center of the three circles.^[4] For illustration, the orange circle in Figure 6 crosses the black given circles at right angles.Inversion in the radical circle leaves the given circles unchanged, but transforms the two conjugate pink solution circles into one another. Under the same inversion, the corresponding points of tangency of the two solution circles are transformed into one another; for illustration, in Figure 6, the two blue points lying on each green line are transformed into one another. Hence, the lines connecting these conjugate tangent points are invariant under the inversion; therefore, they must pass through the center of inversion, which is the radical center (green lines intersecting at the orange dot in Figure 6).

If two of the three given circles do not intersect, a center of inversion can be chosen so that those two given circles becomeconcentric.^[2][12] Under this inversion, the solution circles must fall within theannulus between the two concentric circles. Therefore, they belong to two one-parameter families. In the first family (Figure 7), the solutions donot enclose the inner concentric circle, but rather revolve like ball bearings in the annulus. In the second family (Figure 8), the solution circles enclose the inner concentric circle. There are generally four solutions for each family, yielding eight possible solutions, consistent with thealgebraic solution.

When two of the given circles are concentric, Apollonius's problem can be solved easily using a method ofGauss.^[28] The radii of the three given circles are known, as is the distanced_non from the common concentric center to the non-concentric circle (Figure 7). The solution circle can be determined from its radiusr_s, the angle θ, and the distancesd_s andd_T from its center to the common concentric center and the center of the non-concentric circle, respectively. The radius and distanced_s are known (Figure 7), and the distanced_T =r_s ±r_non, depending on whether the solution circle is internally or externally tangent to the non-concentric circle. Therefore, by thelaw of cosines,

${\displaystyle \cos \theta =\frac{d_{\mathrm{s}}^2+d_{\mathrm{n}\mathrm{o}\mathrm{n}}^2-d_{\mathrm{T}}^2}{2d_{\mathrm{s}}{d}_{\mathrm{n}\mathrm{o}\mathrm{n}}}\equiv C_{\pm }.}$

Here, a new constantC has been defined for brevity, with the subscript indicating whether the solution is externally or internally tangent. A simple trigonometric rearrangement yields the four solutions

${\displaystyle \theta =\pm 2\arctan \left(\sqrt{\frac{1-C}{1+C}}\right).}$

This formula represents four solutions, corresponding to the two choices of the sign of θ, and the two choices forC. The remaining four solutions can be obtained by the same method, using the substitutions forr_s andd_s indicated in Figure 8. Thus, all eight solutions of the general Apollonius problem can be found by this method.

Any initial two disjoint given circles can be rendered concentric as follows. Theradical axis of the two given circles is constructed; choosing two arbitrary pointsP andQ on this radical axis, two circles can be constructed that are centered onP andQ and that intersect the two given circles orthogonally. These two constructed circles intersect each other in two points. Inversion in one such intersection pointF renders the constructed circles into straight lines emanating fromF and the two given circles into concentric circles, with the third given circle becoming another circle (in general). This follows because the system of circles is equivalent to a set ofApollonian circles, forming abipolar coordinate system.

The usefulness ofinversion can be increased significantly by resizing.^[37][38] As noted inViète's reconstruction, the three given circles and the solution circle can be resized in tandem while preserving their tangencies. Thus, the initial Apollonius problem is transformed into another problem that may be easier to solve. For example, the four circles can be resized so that one given circle is shrunk to a point; alternatively, two given circles can often be resized so that they are tangent to one another. Thirdly, given circles that intersect can be resized so that they become non-intersecting, after which themethod for inverting to an annulus can be applied. In all such cases, the solution of the original Apollonius problem is obtained from the solution of the transformed problem by undoing the resizing and inversion.

In the first approach, the given circles are shrunk or swelled (appropriately to their tangency) until one given circle is shrunk to a pointP.^[37] In that case, Apollonius' problem degenerates to theCCPlimiting case, which is the problem of finding a solution circle tangent to the two remaining given circles that passes through the pointP. Inversion in a circle centered onP transforms the two given circles into new circles, and the solution circle into a line. Therefore, the transformed solution is a line that is tangent to the two transformed given circles. There are four such solution lines, which may be constructed from the external and internalhomothetic centers of the two circles. Re-inversion inP and undoing the resizing transforms such a solution line into the desired solution circle of the original Apollonius problem. All eight general solutions can be obtained by shrinking and swelling the circles according to the differing internal and external tangencies of each solution; however, different given circles may be shrunk to a point for different solutions.

In the second approach, the radii of the given circles are modified appropriately by an amount Δr so that two of them are tangential (touching).^[38] Their point of tangency is chosen as the center ofinversion in a circle that intersects each of the two touching circles in two places. Upon inversion, the touching circles become two parallel lines: Their only point of intersection is sent to infinity under inversion, so they cannot meet. The same inversion transforms the third circle into another circle. The solution of the inverted problem must either be (1) a straight line parallel to the two given parallel lines and tangent to the transformed third given circle; or (2) a circle of constant radius that is tangent to the two given parallel lines and the transformed given circle. Re-inversion and adjusting the radii of all circles by Δr produces a solution circle tangent to the original three circles.

Gergonne's approach is to consider the solution circles in pairs.^[1] Let a pair of solution circles be denoted asC_A andC_B (the pink circles in Figure 6), and let their tangent points with the three given circles be denoted asA₁,A₂,A₃, andB₁,B₂,B₃, respectively. Gergonne's solution aims to locate these six points, and thus solve for the two solution circles.

Gergonne's insight was that if a lineL₁ could be constructed such thatA₁ andB₁ were guaranteed to fall on it, those two points could be identified as the intersection points ofL₁ with the given circleC₁ (Figure 6). The remaining four tangent points would be located similarly, by finding linesL₂ andL₃ that containedA₂ andB₂, andA₃ andB₃, respectively. To construct a line such asL₁, two points must be identified that lie on it; but these points need not be the tangent points. Gergonne was able to identify two other points for each of the three lines. One of the two points has already been identified: theradical centerG lies on all three lines (Figure 6).

To locate a second point on the linesL₁,L₂ andL₃, Gergonne noted areciprocal relationship between those lines and theradical axisR of the solution circles,C_A andC_B. To understand this reciprocal relationship, consider the two tangent lines to the circleC₁ drawn at its tangent pointsA₁ andB₁ with the solution circles; the intersection of these tangent lines is thepole point ofL₁ inC₁. Since the distances from that pole point to the tangent pointsA₁ andB₁ are equal, this pole point must also lie on the radical axisR of the solution circles, by definition (Figure 9). The relationship between pole points and their polar lines is reciprocal; if the pole ofL₁ inC₁ lies onR, the pole ofR inC₁ must conversely lie onL₁. Thus, if we can constructR, we can find its poleP₁ inC₁, giving the needed second point onL₁ (Figure 10).

Gergonne found the radical axisR of the unknown solution circles as follows. Any pair of circles has twocenters of similarity; these two points are the two possible intersections of two tangent lines to the two circles. Therefore, the three given circles have six centers of similarity, two for each distinct pair of given circles. Remarkably, these six points lie on four lines, three points on each line; moreover, each line corresponds to theradical axis of a potential pair of solution circles. To show this, Gergonne considered lines through corresponding points of tangency on two of the given circles, e.g., the line defined byA₁/A₂ and the line defined byB₁/B₂. LetX₃ be a center of similitude for the two circlesC₁ andC₂; then,A₁/A₂ andB₁/B₂are pairs of antihomologous points, and their lines intersect atX₃. It follows, therefore, that the products of distances are equal

${\displaystyle \overline{X_3{A}_1}\cdot \overline{X_3{A}_2}=\overline{X_3{B}_1}\cdot \overline{X_3{B}_2}}$

which implies thatX₃ lies on the radical axis of the two solution circles. The same argument can be applied to the other pairs of circles, so that three centers of similitude for the given three circles must lie on the radical axes of pairs of solution circles.

In summary, the desired lineL₁ is defined by two points: the radical centerG of the three given circles and the pole inC₁ of one of the four lines connecting the homothetic centers. Finding the same pole inC₂ andC₃ givesL₂ andL₃, respectively; thus, all six points can be located, from which one pair of solution circles can be found. Repeating this procedure for the remaining three homothetic-center lines yields six more solutions, giving eight solutions in all. However, if a lineL_k does not intersect its circleC_k for somek, there is no pair of solutions for that homothetic-center line.

The techniques of modernalgebraic geometry, and in particularintersection theory, can be used to solve Apollonius's problem. In this approach, the problem is reinterpreted as a statement about circles in thecomplex projective plane. Solutions involvingcomplex numbers are allowed and degenerate situations are counted with multiplicity. When this is done, there are always eight solutions to the problem.^[39]

Every quadratic equation inX,Y, andZ determines a unique conic, its vanishing locus. Conversely, every conic in the complex projective plane has an equation, and that equation is unique up to an overall scaling factor (because rescaling an equation does not change its vanishing locus). Therefore, the set of all conics may be parametrized by five-dimensional projective spaceP⁵, where the correspondence is

${\displaystyle \{[X:Y:Z]\in {\mathbf{P}}^2:A{X}^2+BXY+C{Y}^2+DXZ+EYZ+F{Z}^2=0\}\leftrightarrow [A:B:C:D:E:F]\in {\mathbf{P}}^5.}$

Acircle in the complex projective plane is defined to be a conic that passes through the two pointsO₊ = [1 :i : 0] andO₋ = [1 : −i : 0], wherei denotes a square root of−1. The pointsO₊ andO₋ are called thecircular points. Theprojective variety of all circles is the subvariety ofP⁵ consisting of those points which correspond to conics passing through the circular points. Substituting the circular points into the equation for a generic conic yields the two equations

${\displaystyle A+Bi-C=0,}$

${\displaystyle A-Bi-C=0.}$

Taking the sum and difference of these equations shows that it is equivalent to impose the conditions

${\displaystyle A=C}$ and${\displaystyle B=0}$.

Therefore, the variety of all circles is a three-dimensional linear subspace ofP⁵. After rescaling andcompleting the square, these equations also demonstrate that every conic passing through the circular points has an equation of the form

${\displaystyle (X-aZ{)}^2+(Y-bZ{)}^2=r^2{Z}^2,}$

which is the homogenization of the usual equation of a circle in the affine plane. Therefore, studying circles in the above sense is nearly equivalent to studying circles in the conventional sense. The only difference is that the above sense permits degenerate circles which are the union of two lines. The non-degenerate circles are calledsmooth circles, while the degenerate ones are calledsingular circles. There are two types of singular circles. One is the union of the line at infinityZ = 0 with another line in the projective plane (possibly the line at infinity again), and the other is union of two lines in the projective plane, one through each of the two circular points. These are the limits of smooth circles as the radiusr tends to+∞ and0, respectively. In the latter case, no point on either of the two lines has real coordinates except for the origin[0 : 0 : 1].

LetD be a fixed smooth circle. IfC is any other circle, then, by the definition of a circle,C andD intersect at the circular pointsO₊ andO₋. BecauseC andD are conics,Bézout's theorem impliesC andD intersect in four points total, when those points are counted with the properintersection multiplicity. That is, there are four points of intersectionO₊,O₋,P, andQ, but some of these points might collide. Appolonius' problem is concerned with the situation whereP =Q, meaning that the intersection multiplicity at that point is2; ifP is also equal to a circular point, this should be interpreted as the intersection multiplicity being3.

LetZ_D be the variety of circles tangent toD. This variety is a quadric cone in theP³ of all circles. To see this, consider theincidence correspondence

${\displaystyle \mathrm{\Phi }=\{(r,C)\in D\times {\mathbf{P}}^3:C\text{ is tangent to }D\text{ at }r\}.}$

For a curve that is the vanishing locus of a single equationf = 0, the condition that the curve meetsD atr with multiplicitym means that theTaylor series expansion off|_D vanishes to orderm atr; it is thereforem linear conditions on the coefficients off. This shows that, for eachr, the fiber ofΦ overr is aP¹ cut out by two linear equations in the space of circles. Consequently,Φ is irreducible of dimension2. Since it is possible to exhibit a circle that is tangent toD at only a single point, a generic element ofZ_D must be tangent at only a single point. Therefore, the projectionΦ →P² sending(r,C) toC is abirational morphism. It follows that the image ofΦ, which isZ_D, is also irreducible and two dimensional.

To determine the shape ofZ_D, fix two distinct circlesC₀ andC_∞, not necessarily tangent toD. These two circles determine apencil, meaning a lineL in theP³ of circles. If the equations ofC₀ andC_∞ aref andg, respectively, then the points onL correspond to the circles whose equations areSf +Tg, where[S :T] is a point ofP¹. The points whereL meetsZ_D are precisely the circles in the pencil that are tangent toD.

There are two possibilities for the number of points of intersections. One is that eitherf org, sayf, is the equation forD. In this case,L is a line throughD. IfC_∞ is tangent toD, then so is every circle in the pencil, and thereforeL is contained inZ_D. The other possibility is that neitherf norg is the equation forD. In this case, the function(f /g)|_D is a quotient of quadratics, neither of which vanishes identically. Therefore, it vanishes at two points and haspoles at two points. These are the points inC₀ ∩D andC_∞ ∩D, respectively, counted with multiplicity and with the circular points deducted. The rational function determines a morphismD →P¹ of degree two. The fiber over[S :T] ∈P¹ is the set of pointsP for whichf(P)T =g(P)S. These are precisely the points at which the circle whose equation isTf −Sg meetsD. Thebranch points of this morphism are the circles tangent toD. By theRiemann–Hurwitz formula, there are precisely two branch points, and thereforeL meetsZ_D in two points. Together, these two possibilities for the intersection ofL andZ_D demonstrate thatZ_D is a quadric cone. All such cones inP³ are the same up to a change of coordinates, so this completely determines the shape ofZ_D.

To conclude the argument, letD₁,D₂, andD₃ be three circles. If the intersectionZ_D1 ∩Z_D2 ∩Z_D3 is finite, then it has degree2³ = 8, and therefore there are eight solutions to the problem of Apollonius, counted with multiplicity. To prove that the intersection is generically finite, consider the incidence correspondence

${\displaystyle \mathrm{\Psi }=\{(D_1,D_2,D_3,C)\in ({\mathbf{P}}^3{)}^4:C\text{ is tangent to all }{D}_i\}.}$

There is a morphism which projectsΨ onto its final factor ofP³. The fiber overC isZ_C³. This has dimension6, soΨ has dimension9. Because(P³)³ also has dimension9, the generic fiber of the projection fromΨ to the first three factors cannot have positive dimension. This proves that generically, there are eight solutions counted with multiplicity. Since it is possible to exhibit a configuration where the eight solutions are distinct, the generic configuration must have all eight solutions distinct.

In the generic problem with eight solution circles, The reciprocals of the radii of four of the solution circles sum to the same value as do the reciprocals of the radii of the other four solution circles^[40]

Apollonius problem is to construct one or more circles tangent to three given objects in a plane, which may be circles, points, or lines. This gives rise to ten types of Apollonius' problem, one corresponding to each combination of circles, lines and points, which may be labeled with three letters, eitherC,L, orP, to denote whether the given elements are a circle, line or point, respectively (Table 1).^[32] As an example, the type of Apollonius problem with a given circle, line, and point is denoted asCLP.

Some of thesespecial cases are much easier to solve than the general case of three given circles. The two simplest cases are the problems of drawing a circle through three given points (PPP) or tangent to three lines (LLL), which were solved first byEuclid in hisElements. For example, thePPP problem can be solved as follows. The center of the solution circle is equally distant from all three points, and therefore must lie on theperpendicular bisector line of any two. Hence, the center is the point of intersection of any two perpendicular bisectors. Similarly, in theLLL case, the center must lie on a line bisecting the angle at the three intersection points between the three given lines; hence, the center lies at the intersection point of two such angle bisectors. Since there are two such bisectors at every intersection point of the three given lines, there are four solutions to the generalLLL problem (theincircle and excircles of the triangle formed by the three lines).

Points and lines may be viewed as special cases of circles; a point can be considered as a circle of infinitely small radius, and a line may be thought of an infinitely large circle whose center is also at infinity. From this perspective, the general Apollonius problem is that of constructing circles tangent to three given circles. The nine other cases involving points and lines may be viewed aslimiting cases of the general problem.^[32][12] These limiting cases often have fewer solutions than the general problem; for example, the replacement of a given circle by a given point halves the number of solutions, since a point can be construed as an infinitesimal circle that is either internally or externally tangent.

Table 1: Ten Types of Apollonius' Problem

Index	Code	Given Elements	Number of solutions (in general)	Example (solution in pink; given objects in black)
1	PPP	three points	1
2	LPP	one line and two points	2
3	LLP	two lines and a point	2
4	CPP	one circle and two points	2
5	LLL	three lines	4
6	CLP	one circle, one line, and a point	4
7	CCP	two circles and a point	4
8	CLL	one circle and two lines	8
9	CCL	two circles and a line	8
10	CCC	three circles (the classic problem)	8

The problem of counting the number of solutions to different types of Apollonius' problem belongs to the field ofenumerative geometry.^[12][41] The general number of solutions for each of the ten types of Apollonius' problem is given in Table 1 above. However, special arrangements of the given elements may change the number of solutions. For illustration, Apollonius' problem has no solution if one circle separates the two (Figure 11); to touch both the solid given circles, the solution circle would have to cross the dashed given circle; but that it cannot do, if it is to touch the dashed circle tangentially. Conversely, if three given circles are all tangent at the same point, thenany circle tangent at the same point is a solution; such Apollonius problems have an infinite number of solutions. If any of the given circles are identical, there is likewise an infinity of solutions. If only two given circles are identical, there are only two distinct given circles; the centers of the solution circles form ahyperbola, as used inone solution to Apollonius' problem.

An exhaustive enumeration of the number of solutions for all possible configurations of three given circles, points or lines was first undertaken by Muirhead in 1896,^[42] although earlier work had been done by Stoll^[43] and Study.^[44] However, Muirhead's work was incomplete; it was extended in 1974^[45] and a definitive enumeration, with 33 distinct cases, was published in 1983.^[12] Although solutions to Apollonius' problem generally occur in pairs related byinversion, an odd number of solutions is possible in some cases, e.g., the single solution forPPP, or when one or three of the given circles are themselves solutions. (An example of the latter is given in thesection onDescartes' theorem.) However, there are no Apollonius problems with seven solutions.^[34][43] Alternative solutions based on thegeometry of circles and spheres have been developed and used in higher dimensions.^[26][35]

If the three given circles are mutually tangent, Apollonius' problem has five solutions. Three solutions are the given circles themselves, since each is tangent to itself and to the other two given circles. The remaining two solutions (shown in red in Figure 12) correspond to theinscribed andcircumscribed circles, and are calledSoddy's circles.^[46] This special case of Apollonius' problem is also known as thefour coins problem.^[47] The three given circles of this Apollonius problem form aSteiner chain tangent to the two Soddy's circles.

Either Soddy circle, when taken together with the three given circles, produces a set of four circles that are mutually tangent at six points. The radii of these four circles are related by an equation known asDescartes' theorem. In a 1643 letter to PrincessElizabeth of Bohemia,^[48]René Descartes showed that

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

wherek_s = 1/r_s andr_s are thecurvature and radius of the solution circle, respectively, and similarly for the curvaturesk₁,k₂ andk₃ and radiir₁,r₂ andr₃ of the three given circles. For every set of four mutually tangent circles, there is a second set of four mutually tangent circles that are tangent at the same six points.^[2][49]

Descartes' theorem was rediscovered independently in 1826 byJakob Steiner,^[50] in 1842 by Philip Beecroft,^[2][49] and again in 1936 byFrederick Soddy.^[51] Soddy published his findings in the scientific journalNature as a poem,The Kiss Precise, of which the first two stanzas are reproduced below. The first stanza describes Soddy's circles, whereas the second stanza gives Descartes' theorem. In Soddy's poem, two circles are said to "kiss" if they are tangent, whereas the term "bend" refers to the curvaturek of the circle.

Sundry extensions of Descartes' theorem have been derived byDaniel Pedoe.^[52]

Apollonius' problem can be extended to construct all the circles that intersect three given circles at a precise angle θ, or at three specified crossing angles θ₁, θ₂ and θ₃;^[50] the ordinary Apollonius' problem corresponds to a special case in which the crossing angle is zero for all three given circles. Another generalization is thedual of the first extension, namely, to construct circles with three specified tangential distances from the three given circles.^[26]

Apollonius' problem can be extended from the plane to thesphere and otherquadratic surfaces. For the sphere, the problem is to construct all the circles (the boundaries ofspherical caps) that are tangent to three given circles on the sphere.^[24][53][54] This spherical problem can be rendered into a corresponding planar problem usingstereographic projection. Once the solutions to the planar problem have been constructed, the corresponding solutions to the spherical problem can be determined by inverting the stereographic projection. Even more generally, one can consider the problem of four tangent curves that result from the intersections of an arbitrary quadratic surface and four planes, a problem first considered byCharles Dupin.^[9]

By solving Apollonius' problem repeatedly to find the inscribed circle, the interstices between mutually tangential circles can be filled arbitrarily finely, forming anApollonian gasket, also known as aLeibniz packing or anApollonian packing.^[55] This gasket is afractal, being self-similar and having adimensiond that is not known exactly but is roughly 1.3,^[56] which is higher than that of aregular (orrectifiable) curve (d = 1) but less than that of a plane (d = 2). The Apollonian gasket was first described byGottfried Leibniz in the 17th century, and is a curved precursor of the 20th-centurySierpiński triangle.^[57] The Apollonian gasket also has deep connections to other fields of mathematics; for example, it is the limit set ofKleinian groups.^[58]

The configuration of a circle tangent tofour circles in the plane has special properties, which have been elucidated by Larmor (1891)^[59] and Lachlan (1893).^[60] Such a configuration is also the basis forCasey's theorem,^[17] itself a generalization ofPtolemy's theorem.^[37]

The extension of Apollonius' problem to three dimensions, namely, the problem of finding a fifth sphere that is tangent to four given spheres, can be solved by analogous methods.^[9] For example, the given and solution spheres can be resized so that one given sphere is shrunk to point while maintaining tangency.^[38] Inversion in this point reduces Apollonius' problem to finding a plane that is tangent to three given spheres. There are in general eight such planes, which become the solutions to the original problem by reversing the inversion and the resizing. This problem was first considered byPierre de Fermat,^[61] and many alternative solution methods have been developed over the centuries.^[62]

Apollonius' problem can even be extended tod dimensions, to construct thehyperspheres tangent to a given set ofd + 1 hyperspheres.^[41] Following the publication ofFrederick Soddy's re-derivation of theDescartes' theorem in 1936, several people solved (independently) the mutually tangent case corresponding to Soddy's circles ind dimensions.^[63]

The principal application of Apollonius' problem, as formulated by Isaac Newton, ishyperbolic trilateration, which seeks to determine a position from thedifferences in distances to at least three points.^[8] For example, a ship may seek to determine its position from the differences in arrival times of signals from three synchronized transmitters. Solutions to Apollonius' problem were used inWorld War I to determine the location of an artillery piece from the time a gunshot was heard at three different positions,^[9] and hyperbolic trilateration is the principle used by theDecca Navigator System andLORAN.^[7] Similarly, the location of an aircraft may be determined from the difference in arrival times of itstransponder signal at four receiving stations. Thismultilateration problem is equivalent to the three-dimensional generalization of Apollonius' problem and applies toglobal navigation satellite systems (seeGPS#Geometric interpretation).^[31] It is also used to determine the position of calling animals (such as birds and whales), although Apollonius' problem does not pertain if thespeed of sound varies with direction (i.e., thetransmission medium notisotropic).^[64]

Apollonius' problem has other applications. In Book 1, Proposition 21 in hisPrincipia,Isaac Newton used his solution of Apollonius' problem to construct an orbit incelestial mechanics from the center of attraction and observations of tangent lines to the orbit corresponding to instantaneousvelocity.^[9] The special case of the problem of Apollonius when all three circles are tangent is used in theHardy–Littlewood circle method ofanalytic number theory to constructHans Rademacher's contour for complex integration, given by the boundaries of aninfinite set ofFord circles each of which touches several others.^[65] Finally, Apollonius' problem has been applied to some types ofpacking problems, which arise in disparate fields such as theerror-correcting codes used onDVDs and the design of pharmaceuticals that bind in a particularenzyme of a pathogenicbacterium.^[66]

• Apollonius point
• Apollonius' theorem
• Isodynamic point of a triangle

EnglishWikisource has original text related to this article:

Properties of circles in mutual contact

FrenchWikisource has original text related to this article:

Poncelet's solution to the problem of Apollonius

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• Camerer, JG (1795).Apollonii de Tactionibus, quae supersunt, ac maxime lemmata Pappi, in hos libros Graece nunc primum edita, e codicibus manuscriptis, cum Vietae librorum Apollonii restitutione, adjectis observationibus, computationibus, ac problematis Apolloniani historia (in Latin). Gothae: Ettinger.
• Gisch D, Ribando JM (2004)."Apollonius' Problem: A Study of Solutions and Their Connections"(PDF).American Journal of Undergraduate Research.3:15–25.doi:10.33697/ajur.2004.010. Archived fromthe original(PDF) on 2008-04-15. Retrieved2009-04-16.
• Pappus of Alexandria (1933).Pappus d'Alexandrie: La collection mathématique (in French). Paris: Desclée de Brouwer.OCLC 67245614. Trans., introd., and notes by Paul Ver Eecke.
• Simon, M (1906).Über die Entwicklung der Elementargeometrie im XIX. Jahrhundert (in German). Berlin: Teubner. pp. 97–105.
• Wells, D (1991).The Penguin Dictionary of Curious and Interesting Geometry. New York: Penguin Books. pp. 3–5.ISBN 0-14-011813-6.
• Joseph Howlett (September 26, 2025)."New Math Revives Geometry's Oldest Problems".Quanta Magazine.

Wikimedia Commons has media related toProblem of Apollonius.

• "Ask Dr. Math solution". Mathforum. Retrieved2008-05-05.
• Weisstein, Eric W."Apollonius' problem".MathWorld.
• "Apollonius' Problem".Cut The Knot. Retrieved2008-05-05.
• Kunkel, Paul."Tangent Circles". Whistler Alley. Retrieved2008-05-05.
• Austin, David (March 2006)."When kissing involves trigonometry". Feature Column at the American Mathematical Society website. Retrieved2008-05-05.

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