Problem of Apollonius

Problem of Apollonius

In Euclidean plane geometry, Apollonius' problem is to construct circles that are tangent to three given circles in a plane (Figure 1); two circles are tangent if they touch at a single point. Apollonius of Perga (ca. 262 BC &ndash; ca. 190 BC) posed and solved this famous problem in his work "Επαφαι" ("Tangencies"), which has been lost. A 4th-century report of his results by Pappus of Alexandria has survived, from which François Viète was able to reconstruct Apollonius' solution in the 16th century.

Apollonius' problem has been a challenge throughout its history, with mathematicians such as René Descartes, Leonard Euler, Carl Friedrich Gauss and Joseph Diaz Gergonne contributing solutions. An early solution by Adriaan van Roomen used intersections of hyperbolas. Later work provided solutions using straightedge and compass constructions, algebraic formulas, or additional symmetries. The algebraic solutions have been important in practical applications such as trilateration; a position can be found from differences in distances to three known points.

In general, three given circles have eight different circles that are tangent to them (Figure 2). The solution circles differ from one another in how they enclose (or exclude) the three given circles; each solution corresponds to one way of enclosing or excluding the given circles. However, the problem has many special or limiting cases. Any of the three given circles can be shrunk to zero radius (a point) or expanded to infinite radius (a line). Another important special case is when the three given circles are tangent to each other. In this case, René Descartes found an equation relating the radii of the solution circles and those of the given circles; this equation is now known as Descartes' theorem.

Apollonius' problem can be generalized in several ways. Instead of lying in a plane, the three given circles may lie on a sphere or other quadric surface. Another generalization is to construct circles that cross the three given circles at three specified angles instead of being tangent to them. The three-dimensional analog of Apollonius' problem is to construct a sphere tangent to four given spheres; generalizations to even higher dimensions are also possible. One of the earliest fractals to be described in print was the Apollonian gasket proposed by Gottfried Leibniz, which is based on solving Apollonius' problem iteratively.

tatement of the problem

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.cite book | author = Dörrie H | year = 1965 | chapter = The Tangency Problem of Apollonius | title = 100 Great Problems of Elementary Mathematics: Their History and Solutions | publisher = Dover | location = New York | pages = pp. 154&ndash;160 (§32)] 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 called "Apollonius circles", although the term is also used for other 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 to "intersect" 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 be "tangent"; the intersection point is called a "tangent point" or a "point of tangency". (The word "tangent" derives from the Latin present 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, although two parallel lines can be considered as tangent at a point at infinity.

The solution circle may be either internally or externally tangent to each of the given circles. An "external" tangency is one where the two circles bend away from each other at their point of contact; they lie on opposite sides of the tangent line at that point, and they exclude one another. The distance between their centers equals the sum of their radii. By contrast, an "internal" 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 the "differences" of its distances to three given points equal three known values. Consider a solution circle of radius "r""s" and three given circles of radii "r"1, "r"2 and "r"3. 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 equal "d"1 = "r"1 + "r""s", "d"2 = "r"2 + "r""s" and "d"3 = "r"3 + "r""s", respectively. Therefore, differences in these distances are constants, such as "d"1 − "d"2 = "r"1 − "r"2; they depend only on the known radii of the given circles and not on the radius "r""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 radius "r""s" again cancels out. The re-formulation in terms of center-center distances is useful in the solutions below of Adriaan van Roomen and Isaac Newton, and also in trilateration, which is the task of locating a position from differences in distances to three known points. For example, navigation systems such as GPS and LORAN 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.

History

A rich repertoire of geometrical and algebraic methods have been developed to solve Apollonius' problem,cite journal | author = Althiller-Court N | year = 1961 | title = The problem of Apollonius | journal = The Mathematics Teacher | volume = 54 | pages = 444&ndash;452] cite book | author = Gabriel-Marie F | year = 1912 | title = Exercices de géométrie, comprenant l'esposé des méthodes géométriques et 2000 questions résolues | publisher = Maison A. Mame et Fils | location = Tours | pages = [http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=ACV3924.0001.001;didno=ACV3924.0001.001;view=pdf;seq=00000048 18&ndash;20] , [http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=ACV3924.0001.001;didno=ACV3924.0001.001;view=pdf;seq=00000703 673&ndash;677] | url = http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=ACV3924 fr icon] which has been called "the most famous of all" geometry problems.cite book | author = Coolidge JL | year = 1916 | title = A Treatise on the Circle and the Sphere | publisher = Clarendon Press | location = Oxford | pages = 167&ndash;172] The original approach of Apollonius of Perga has been lost, but reconstructions have been offered by François Viète and others, based on the clues in the description by Pappus.cite book | author = Pappus | year = 1876 | title = Pappi Alexandrini collectionis quae supersunt | editor = F Hultsch | edition = 3 volumes la icon] The first new solution method was published in 1596 by Adriaan van Roomen, who identified the centers of the solution circles as the intersection points of two hyperbolas.cite book | author = van Roomen A | year = 1596 | title = Problema Apolloniacum quo datis tribus circulis, quaeritur quartus eos contingens, antea a...Francisco Vieta...omnibus mathematicis...ad construendum propositum, jam vero per Belgam...constructum | publisher = Würzburg la icon] cite book | author = Newton I | year = 1974 | title = The Mathematical Papers of Isaac Newton, Volume VI: 1684&ndash;1691 | editor = DT Whiteside | publisher = Cambridge University Press | location = Cambridge | isbn = 0-521-08719-8| pages = p. 164] Van Roomen's method was refined in 1687 by Isaac Newton in his "Principia",cite book | author = Newton I | year = 1687 | title = Philosophiæ Naturalis Principia Mathematica | pages = Book I, Section IV, Lemma 16] [cite book | author = Newton I | year = 1974 | title = The Mathematical Papers of Isaac Newton, Volume VI: 1684&ndash;1691 | editor = DT Whiteside | publisher = Cambridge University Press | location = Cambridge | isbn = 0-521-08719-8| pages = pp. 162&ndash;165, 238&ndash;241] and independently by John Casey in 1881.cite book | author = Casey J | year = 1881 | title = A Sequel to Euclid | pages = p. 122]

Although successful in solving Apollonius' problem, van Roomen's method has a drawback. A prized property in classical Euclidean geometry is the ability to solve problems using only a compass and a straightedge. [cite book | author = Courant R, Robbins H | date = 1943 | title = What is Mathematics? An Elementary Approach to Ideas and Methods | publisher = Oxford University Press | location = London | pages = pp. 125&ndash;127, 161&ndash;162] Many constructions are impossible using only these tools, such as dividing an angle in three equal parts. However, many such "impossible" problems can be solved by intersecting curves such as hyperbolas, ellipses and parabolas (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, but Menaechmus showed that the problem can be solved by using the intersections of two parabolas. [cite book |author=Bold B| title = Famous problems of geometry and how to solve them | publisher = Dover Publications | date = 1982 | pages = 29-30 | isbn = 0486242978] 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 friend François Viète, who had urged van Roomen to work on Apollonius' problem in the first place, developed a method that used only compass and straightedge.cite book | author = Viète F | year = 1970 | title = Apollonius Gallus. Seu, Exsuscitata Apolloni Pergæi Περι Επαφων Geometria | edition = photographic reproduction of the "Opera Mathematica" (1646), Schooten FA, editor | publisher = Georg Olms | location = New York | pages = pp. 325&ndash;346 la icon] Prior to Viète's solution, Regiomontanus doubted whether Apollonius' problem could be solved by straightedge and compass.cite book| author = Boyer CB, Merzbach UC | year = 1991 | title = A History of Mathematics | edition= 2nd edition | publisher = John Wiley & Sons, Inc. |isbn=0-471-54397-7| chapter = Apollonius of Perga | pages = p. 322] 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. According to the 4th-century report of Pappus of Alexandria, Apollonius' own book on this problem—entitled Επαφαι ("Tangencies", Latin: "De tactionibus", "De contactibus")—followed a similar progressive approach. Hence, Viète's solution is considered to be a plausible reconstruction of Apollonius' solution, although another reconstruction has been published, independently by three different authors.Simson R (1734) "Mathematical Collection", volume VII, p. 117.
cite book | author = Zeuthen HG | year = 1886 | title = Die Lehre von den Kegelschnitten im Altertum | publisher = Unknown | location = Copenhagen | pages = pp. 381&ndash;383 de icon
cite book | author = Heath TL | title = A History of Greek Mathematics, Volume II: From Aristarchus to Diophantus | publisher = Clarendon Press | location = Oxford | pages = pp. 181&ndash;185, 416&ndash;417]

Several other geometrical solutions to Apollonius' problem were developed in the 19th century. The most notable solutions are those of Jean-Victor Poncelet (1811) [cite journal | author = Poncelet J-V | date = January 1811 | title = Solutions de plusieurs problêmes de géométrie et de mécanique | journal = Correspondance sur l'École Impériale Polytechnique | volume = 2 | issue = 3 | pages = pp. 271&ndash;273 fr icon] and of Joseph Diaz Gergonne (1814).cite journal | author = Gergonne J | date = 1813&ndash;1814 |title = Recherche du cercle qui en touche trois autres sur une sphère | journal = Ann. Math. Pures appl. |volume = 4 fr icon] Whereas Poncelet's proof relies on homothetic centers of circles and the power of a point theorem, Gergonne's method exploits the conjugate relation between lines and their poles in a circle. Methods using circle inversion were pioneered by Julius Petersen in 1879;cite book | author = Petersen J | year = 1879 | title = Methods and Theories for the Solution of Problems of Geometrical Constructions, Applied to 410 Problems | publisher = Sampson Low, Marston, Searle & Rivington | location = London | pages = pp. 94&ndash;95 (Example 403)] one example is the annular solution method of HSM Coxeter.cite journal | author = Coxeter HSM | year = 1968 | title = The Problem of Apollonius | journal = The American Mathematical Monthly | volume = 75 | pages = pp. 5&ndash;15 | doi = 10.2307/2315097] Another approach uses Lie sphere geometry, which was developed by Sophus Lie.

Algebraic solutions to Apollonius' problem were pioneered in the 17th century by René Descartes and Princess Elisabeth of Bohemia, although their solutions were rather complex. Practical algebraic methods were developed in the late 18th and 19th centuries by several mathematicians, including Leonhard Euler, [cite journal | author = Euler L | year = 1790 | title = Solutio facilis problematis, quo quaeritur circulus, qui datos tres circulos tangat | journal = Nova Acta Academiae Scientarum Imperialis Petropolitinae | volume = 6 | pages = 95&ndash;101 | url = http://www.math.dartmouth.edu/~euler/docs/originals/E648.pdf la icon Reprinted in Euler's "Opera Omnia", series 1, volume 26, pp. 270&ndash;275.] Nicolas Fuss, Carl Friedrich Gauss,cite book | author = Gauss CF | year = 1873 | title = Werke, 4. Band | publisher = Königlichen Gesellschaft der Wissenschaften | location = Göttingen | edition = reprinted in 1973 by Georg Olms Verlag (Hildesheim) | pages = 399&ndash;400 | isbn = 3-487-04636-9 de icon] Lazare Carnot,cite book | author = Carnot L | year = 1801 | title = De la corrélation dans les figures de géométrie | publisher = Unknown publisher | location = Paris | pages = No. 158&ndash;159 fr icon
cite book | author = Carnot L | year = 1803 | title = Géométrie de position | publisher = Unknown publisher | location = Paris | pages = p. 390, &sect;334 fr icon] and Augustin Louis Cauchy. [cite journal | author = Cauchy AL | date = July 1806 | title = Du cercle tangent à trois cercles donnés | journal = Correspondance sur l'École Polytechnique | volume = 1 | issue = 6 | pages = pp. 193&ndash;195 fr icon]

olution methods

Intersecting hyperbolas

The solution of Adriaan van Roomen (1596) is based on the intersection of two hyperbolas. Let the given circles be denoted as "C"1, "C"2 and "C"3. Van Roomen solved the general problem by solving a simpler problem, that of finding the circles that are tangent to "two" given circles, such as "C"1 and "C"2. He noted that the center of a circle tangent to both given circles must lie on a hyperbola 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 as "r""s", "r""1" and "r""2", respectively (Figure 3). The distance "d"1 between the centers of the solution circle and "C"1 is either "r""s" + "r""1" or "r""s" − "r""1", depending on whether these circles are chosen to be externally or internally tangent, respectively. Similarly, the distance "d"2 between the centers of the solution circle and "C"2 is either "r""s" + "r""2" or "r""s" − "r""2", again depending on their chosen tangency. Thus, the difference "d"1 − "d"2 between these distances is always a constant that is independent of "r""s". This property, of having a fixed difference between the distances to the foci, 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 circles "C"2 and "C"3, where the internal or external tangency of the solution and "C"2 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. Newton formulates Apollonius' problem as a problem in trilateration: to locate a point Z from three given points A, B and C, such that the differences in distances from Z to the three given points have known values. These four points correspond to the center of the solution circle (Z) and the centers of the three given circles (A, B and C).

Instead of solving for the two hyperbolas, Newton constructs their directrix lines instead. For any hyperbola, the ratio of distances from a point Z to a focus A and to the directrix is a fixed constant called the eccentricity. The two directrices intersect at a point T, and from their two known distance ratios, Newton constructs a line passing through T on which Z 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 a circle can be defined as the set of points that have a given ratio of distances to two fixed points. (As an aside, this definition is the basis of bipolar coordinates.) Thus, the solutions to Apollonius' problem are the intersections of a line with a circle.

Viète's reconstruction

As described below, 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 as CCP. 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.

Viète began by solving the PPP case (three points) following the method of Euclid in his "Elements". From this, he derived a lemma corresponding to the power of a point theorem, which he used to solve the LPP case (a line and two points). Following Euclid a second time, Viète solved the LLL case (three lines) using the angle 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 the LLP 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 Δ"r", the radius of its internally tangent given circles must be likewise changed by Δ"r", whereas the radius of its externally tangent given circles must be changed by −Δ"r". 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 the CLL case (a circle and two lines) by shrinking the circle into a point, rendering it a LLP case. He then solved the CLP case (a circle, a line and a point) using three lemmas. Again shrinking one circle to a point, Viète transformed the CCL case into a CLP case. He then solved the CPP case (a circle and two points) and the CCP case (two circles and a point), the latter case by two lemmas. Finally, Viète solved the general CCC case (three circles) by shrinking one circle to a point, rendering it a CCP case.

Algebraic solutions

Apollonius' problem can be framed as a system of three equations for the center and radius of the solution circle.cite journal | author = Coaklay GW | year = 1860 | title = Analytical Solutions of the Ten Problems in the Tangencies of Circles; and also of the Fifteen Problems in the Tangencies of Spheres | journal = The Mathematical Monthly | volume = 2 | pages = 116&ndash;126] 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 ("x"1, "y"1), ("x"2, "y"2) and ("x"3, "y"3), 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 as "r"1, "r"2, "r"3 and "r""s", respectively. The requirement that a solution circle must exactly touch each of the three given circles can be expressed as three coupled quadratic equations for "x""s", "y""s" and "r""s":

:$left\left( x_\left\{s\right\} - x_\left\{1\right\} ight\right)^\left\{2\right\} +left\left( y_\left\{s\right\} - y_\left\{1\right\} ight\right)^\left\{2\right\} = left\left( r_\left\{s\right\} - s_\left\{1\right\} r_\left\{1\right\} ight\right)^\left\{2\right\}$

:$left\left( x_\left\{s\right\} - x_\left\{2\right\} ight\right)^\left\{2\right\} +left\left( y_\left\{s\right\} - y_\left\{2\right\} ight\right)^\left\{2\right\} = left\left( r_\left\{s\right\} - s_\left\{2\right\} r_\left\{2\right\} ight\right)^\left\{2\right\}$

:$left\left( x_\left\{s\right\} - x_\left\{3\right\} ight\right)^\left\{2\right\} +left\left( y_\left\{s\right\} - y_\left\{3\right\} ight\right)^\left\{2\right\} = left\left( r_\left\{s\right\} - s_\left\{3\right\} r_\left\{3\right\} ight\right)^\left\{2\right\}.$

The three numbers "s"1, "s"2 and "s"3 on the right-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 of resultants. When multiplied out, all three equations have "x""s"2 + "y""s"2 on the left-hand side, and "r""s"2 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 coordinates "x""s" and "y""s"

:$x_\left\{s\right\} = M + N r_\left\{s\right\}$

:$y_\left\{s\right\} = P + Q r_\left\{s\right\}$

where "M", "N", "P" and "Q" 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 for "r""s", which can be solved by the quadratic formula. Substitution of the numerical value of "r""s" into the linear formulae yields the corresponding values of "x""s" and "y""s".

The signs "s"1, "s"2 and "s"3 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 for "r""s" is quadratic. 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 signs "s""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 different real numbers, two identical real numbers (i.e., a degenerate double root), or a pair of complex conjugate roots. The first case corresponds to the usual situation; each pair of roots corresponds to a pair of solutions that are related by circle 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 do 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. Interestingly, Apollonius' problem cannot have 7 solutions, although it may have any other number of solutions from zero to eight.cite journal | author = Pedoe D | date = 1970 | title = The missing seventh circle | journal = Elemente der Mathematik | volume = 25 | pages = 14&ndash;15]

Lie sphere geometry

The same algebraic equations can be derived in the context of Lie sphere geometry.cite journal | author = Zlobec BJ, Kosta NM | year = 2001 | title = Configurations of Cycles and the Apollonius Problem | journal = Rocky Mountain Journal of Mathematics | volume = 31 | pages = 725&ndash;744 | doi = 10.1216/rmjm/1020171586] That geometry represents circles, lines and points in a unified way, as a five-dimensional vector "X" = ("v", "c""x", "c""y", "w", "sr"), where c = ("c""x", "c""y") is the center of the circle, and "r" is its (non-negative) radius. If "r" is not zero, the sign "s" may be positive or negative; for visualization, "s" represents the orientation of the circle, with counterclockwise circles having a positive "s" and clockwise circles having a negative "s". The parameter "w" is zero for a straight line, and one otherwise.

In this five-dimensional world, there is a bilinear product similar to the dot product:

:$left\left( X_\left\{1\right\} | X_\left\{2\right\} ight\right) :=v_\left\{1\right\} w_\left\{2\right\} + v_\left\{2\right\} w_\left\{1\right\} + mathbf\left\{c\right\} cdot mathbf\left\{c\right\} - s_\left\{1\right\} s_\left\{2\right\} r_\left\{1\right\} r_\left\{2\right\}.$

The Lie quadric is defined as those vectors whose product with themselves (their square norm) is zero, ("X"|"X") = 0. Let "X"1 and "X"2 be two vectors belonging to this quadric; the norm of their difference equals

:$left\left( X_\left\{1\right\} - X_\left\{2\right\} | X_\left\{1\right\} - X_\left\{2\right\} ight\right) = 2 left\left( v_\left\{1\right\} - v_\left\{2\right\} ight\right) left\left( w_\left\{1\right\} - w_\left\{2\right\} ight\right) +left\left( mathbf\left\{c\right\}_\left\{1\right\} - mathbf\left\{c\right\}_\left\{2\right\} ight\right) cdot left\left( mathbf\left\{c\right\}_\left\{1\right\} - mathbf\left\{c\right\}_\left\{2\right\} ight\right)- left\left( s_\left\{1\right\} r_\left\{1\right\} - s_\left\{2\right\} r_\left\{2\right\} ight\right)^\left\{2\right\}.$

The product distributes over addition and subtraction (more precisely, it is bilinear):

:$left\left( X_\left\{1\right\} - X_\left\{2\right\} | X_\left\{1\right\} - X_\left\{2\right\} ight\right) = left\left( X_\left\{1\right\} | X_\left\{1\right\} ight\right) - 2 left\left( X_\left\{1\right\} | X_\left\{2\right\} ight\right) + left\left( X_\left\{2\right\} | X_\left\{2\right\} ight\right).$

Since ("X"1|"X"1) = ("X"2|"X"2) = 0 (both belong to the Lie quadric) and since "w"1 = "w"2 = 1 for circles, the product of any two such vectors on the quadric equals

:$- 2 left\left( X_\left\{1\right\} | X_\left\{2\right\} ight\right) = left| mathbf\left\{c\right\}_\left\{1\right\} - mathbf\left\{c\right\}_\left\{2\right\} ight|^\left\{2\right\}- left\left( s_\left\{1\right\} r_\left\{1\right\} - s_\left\{2\right\} r_\left\{2\right\} ight\right)^\left\{2\right\}.$

where the vertical bars sandwiching c1c2 represent the length of that difference vector, i.e., the Euclidean norm. This formula shows that if two quadric vectors "X"1 and "X"2 are orthogonal (perpendicular) to one another—that is, if ("X"1|"X"2) = 0—then their corresponding circles are tangent. For if the two signs "s"1 and "s"2 are the same (i.e. the circles have the same "orientation"), the circles are internally tangent; the distance between their centers equals the "difference" in the radii

:$left| mathbf\left\{c\right\}_\left\{1\right\} - mathbf\left\{c\right\}_\left\{2\right\} ight|^\left\{2\right\} =left\left( r_\left\{1\right\} - r_\left\{2\right\} ight\right)^\left\{2\right\}.$

Conversely, if the two signs "s"1 and "s"2 are different (i.e. the circles have opposite "orientations"), the circles are externally tangent; the distance between their centers equals the "sum" of the radii

:$left| mathbf\left\{c\right\}_\left\{1\right\} - mathbf\left\{c\right\}_\left\{2\right\} ight|^\left\{2\right\}= left\left( r_\left\{1\right\} + r_\left\{2\right\} ight\right)^\left\{2\right\}.$

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 vectors "X"sol that belong to the Lie quadric and are also orthogonal (perpendicular) to the vectors "X"1, "X"2 and "X"3 corresponding to the given circles.

:$left\left( X_\left\{mathrm\left\{sol | X_\left\{mathrm\left\{sol ight\right) = left\left( X_\left\{mathrm\left\{sol | X_\left\{1\right\} ight\right) = left\left( X_\left\{mathrm\left\{sol | X_\left\{2\right\} ight\right) = left\left( X_\left\{mathrm\left\{sol | X_\left\{3\right\} ight\right) = 0$

The advantage of this re-statement is that one can exploit theorems from linear algebra on the maximum number of linearly independent, simultaneously perpendicular vectors. This gives another way to calculate the maximum number of solutions and extend the theorem to higher dimensional spaces.

Inversive methods

A natural setting for problem of Apollonius is inversive geometry. 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 the Euclidean plane isometries; however, they do not simplify the problem, since they merely shift, rotate, and mirror the original problem.

Inversion in a circle with center O and radius "R" consists of the following operation (Figure 5): every point P is mapped into a new point P' such that O, P, and P' are collinear, and the product of the distances of P and P' to the center O equal the radius "R" squared

:$overline\left\{mathbf\left\{OP cdot overline\left\{mathbf\left\{OP^\left\{prime\right\} = R^\left\{2\right\}.$

Thus, if P lies outside the circle, then P' lies within, and vice versa. When P is the same as O, the inversion is said to send P to infinity. (In complex analysis, "infinity" is defined in terms of the Riemann 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 of Möbius transformations on the Riemann sphere. The planar Apollonius problem can be transferred to the sphere by a inverse 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.cite book | author = Salmon G | year = 1879 | title = A Treatise on Conic Sections, Containing an Account of Some of the Most Important Modern Algebraic and Geometric Methods | publisher = Longmans, Green and Co. | location = London | pages = 110&ndash;115, 291&ndash;292]

Pairs of solutions by inversion

Solutions to Apollonius' problem generally occur in pairs; for each solution circle, there is a conjugate solution circle (Figure 6). 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 by inversion, by the following argument.

In general, any three distinct circles have a unique circle—the radical circle—that intersects all of them perpendicularly; the center of that circle is the radical center of the three circles.cite book | title = Geometry Revisited | author = Coxeter HSM, Greitzer SL| year = 1967 | publisher = MAA | location = Washington | isbn = 978-0883856192] 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).

Inversion to an annulus

If two of the three given circles do not intersect, a center of inversion can be chosen so that those two given circles become concentric. Under this inversion, the solution circles must fall within the annulus between the two concentric circles. Therefore, they belong to two one-parameter families. In the first family (Figure 7), the solutions do "not" 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 the algebraic solution.

When two of the given circles are concentric, Apollonius' problem can be solved easily using a method of Gauss. The radii of the three given circles are known, as is the distance "d"non from the common concentric center to the non-concentric circle (Figure 7). The solution circle can be determined from its radius "r"s, the angle θ, and the distances "d"s and "d"T from its center to the common concentric center and the center of the non-concentric circle, respectively. The radius and distance "d"s are known (Figure 7), and the distance "d"T = "r"s ± "r"non, depending on whether the solution circle is internally or externally tangent to the non-concentric circle. Therefore, by the law of cosines,

:$cos heta = frac\left\{d_\left\{mathrm\left\{s^\left\{2\right\} + d_\left\{mathrm\left\{non^\left\{2\right\} - d_\left\{mathrm\left\{T^\left\{2\left\{2 d_\left\{mathrm\left\{s d_\left\{mathrm\left\{non\right\} equiv C_\left\{pm\right\}.$

Here, a new constant "C" 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

:$heta = pm 2 mathrm\left\{atan\right\}left\left( sqrt\left\{frac\left\{1 - C\right\}\left\{1 + C ight\right).$

This formula represents four solutions, corresponding to the two choices of the sign of θ, and the two choices for "C". The remaining four solutions can be obtained by the same method, using the substitutions for "r"s and "d"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. The radical axis of the two given circles is constructed; choosing two arbitrary points P and Q on this radical axis, two circles can be constructed that are centered on P and Q and that intersect the two given circles orthogonally. These two constructed circles intersect each other in two points. Inversion in one such intersection point F renders the constructed circles into straight lines emanating from F 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 of Apollonian circles, forming a bipolar coordinate system.

Resizing and inversion

The usefulness of inversion can be increased significantly by resizing. As noted in Viète'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 the method 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.

hrinking one given circle to a point

In the first approach, the given circles are shrunk or swelled (appropriately to their tangency) until one given circle is shrunk to a point P.cite book | author = Johnson RA | year = 1960 | title = Advanced Euclidean Geometry: An Elementary treatise on the geometry of the Triangle and the Circle | edition = reprint of 1929 edition by Houghton Mifflin | publisher = Dover Publications | location = New York | pages = pp. 117&ndash;121 (Apollonius' problem), 121&ndash;128 (Casey's and Hart's theorems) | isbn = 978-0486462370] In that case, Apollonius' problem degenerates to the CCP , which is the problem of finding a solution circle tangent to the two remaining given circles that passes through the point P. Inversion in a circle centered on P 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 internal homothetic centers of the two circles. Re-inversion in P 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.

Resizing two given circles to tangency

In the second approach, the radii of the given circles are modified appropriately by an amount Δ"r" so that two of them are tangential (touching).cite book | author = Ogilvy CS |year = 1990 | title = Excursions in Geometry | publisher = Dover | isbn = 0-486-26530-7 | pages = pp. 48&ndash;51 (Apollonius' problem), 60 (extension to tangent spheres)] Their point of tangency is chosen as the center of inversion 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 Δ"r" produces a solution circle tangent to the original three circles.

Gergonne's solution

Gergonne's approach is to consider the solution circles in pairs. Let a pair of solution circles be denoted as "C"A and "C"B (the pink and black circles in Figure 6), and let their tangent points with the three given circles be denoted as A1, A2, A3, and B1, 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 line "L"1 could be constructed such that A1 and B1 were guaranteed to fall on it, those two points could be identified as the intersection points of "L"1 with the given circle "C"1 (Figure 6). The remaining four tangent points would be located similarly, by finding lines "L"2 and "L"3 that contained A2 and B2, and A3 and B3, respectively. To construct a line such as "L"1, 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: the radical center G lies on all three lines (Figure 6).

To locate a second point on the lines "L"1, "L"2 and "L"3, Gergonne noted a reciprocal relationship between those lines and the radical axis "R" of the solution circles, "C"A and "C"B. To understand this reciprocal relationship, consider the two tangent lines to the circle "C"1 drawn at its tangent points A1 and B1 with the solution circles; the intersection of these tangent lines is the pole point of "L"1 in "C"1. Since the distances from that pole point to the tangent points A1 and B1 are equal, this pole point must also lie on the radical axis "R" of the solution circles, by definition (Figure 9). The relationship between pole points and their polar lines is reciprocal; if the pole of "L"1 in "C"1 lies on "R", the pole of "R" in "C"1 must conversely lie on "L"1. Thus, if we can construct "R", we can find its pole P1 in "C"1, giving the needed second point on "L"1 (Figure 10).

Gergonne found the radical axis "R" of the unknown solution circles as follows. Any pair of circles has two centers 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 the radical 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 by A1/A2 and the line defined by B1/B2. Let X3 be a center of similitude for the two circles "C"1 and "C"2; then, A1/A2 and B1/B2 are pairs of antihomologous points, and their lines intersect at X3. It follows, therefore, that the products of distances are equal

:$overline\left\{X_\left\{3\right\}A_\left\{1 cdot overline\left\{X_\left\{3\right\}A_\left\{2 = overline\left\{X_\left\{3\right\}B_\left\{1 cdot overline\left\{X_\left\{3\right\}B_\left\{2$

which implies that X3 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 line "L"1 is defined by two points: the radical center G of the three given circles and the pole in "C"1 of one of the four lines connecting the homothetic centers. Finding the same pole in "C"2 and "C"3 gives "L"2 and "L"3, 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 line "L""k" does not intersect its circle "C""k" for some "k", there is no pair of solutions for that homothetic-center line.

pecial cases

Ten combinations of points, circles and lines

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, either C, L or P, to denote whether the given elements are a circle, line or point, respectively (Table 1). As an example, the type of Apollonius problem with a given circle, line and point is denoted as CLP.

Some of these special 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 by Euclid in his "Elements". For example, the PPP problem can be solved as follows. The center of the solution circle is equally distant from all three points, and therefore must lie on the perpendicular bisector line of any two. Hence, the center is the point of intersection of any two perpendicular bisectors. Similarly, in the LLL 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 general LLL problem.

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 as limiting cases of the general problem.cite book | author = Altshiller-Court N | year = 1952 | title = College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle | edition = 2nd edition, revised and enlarged | publisher = Barnes and Noble | location = New York | pages = 222&ndash;227 | isbn = 978-0486458052
cite book | author = Hartshorne, Robin | year = 2000 | title = Geometry: Euclid and Beyond | publisher = Springer Verlag | location = New York |isbn= 978-0387986500 |pages= pp. 346&ndash;355, 496, 499
cite book | author = Rouché, Eugène |coauthors= Ch de Comberousse |year = 1883 | title = Traité de géométrie | edition = 5th edition, revised and augmented |publisher= Gauthier-Villars |location = Paris | pages = pp. 252&ndash;256 |oclc= 252013267 fr icon] 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.

Number of solutions

The problem of counting the number of solutions to different types of Apollonius' problem belongs to the field of enumerative geometry.cite journal|journal=Acta Mathematica Universitatis Comenianae|volume=68|issue=1|year=1999|pages=37&ndash;47|title=Apollonius' contact problem in "n"-space in view of enumerative geometry|author = Dreschler K, Sterz U|url=http://www.emis.de/journals/AMUC/_vol-68/_no_1/_drechsl/drechsle.html] 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, then "any" 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 a hyperbola, as used in one 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,cite journal| author = Muirhead RF | year = 1896 | title = On the Number and nature of the Solutions of the Apollonian Contact Problem | journal = Proceedings of the Edinburgh Mathematical Society | volume = 14 | pages = 135&ndash;147, attached figures 44&ndash;114] although earlier work had been done by Stollcite journal | author = Stoll V | year = 1876 | title = Zum Problem des Apollonius | journal = Mathematische Annalen | volume = 6 | pages = 613&ndash;632 | doi = 10.1007/BF01443201 de icon] and Study.cite journal | author = Study E | year = 1897 | title = Das Apollonische Problem | journal = Mathematische Annalen | volume = 49 | pages = 497&ndash;542 | doi = 10.1007/BF01444366 de icon] However, Muirhead's work was incomplete; it was extended in 1974cite journal | author = Fitz-Gerald JM | year = 1974 | title = A Note on a Problem of Apollonius | journal = Journal of Geometry | volume = 5 | pages = 15&ndash;26 | doi = 10.1007/BF01954533] and a definitive enumeration was published in 1983.cite journal | author = Bruen A, Fisher JC, Wilker JB | year = 1983 | title = Apollonius by Inversion | journal = Mathematics Magazine | volume = 56 | pages = 97&ndash;103] Although solutions to Apollonius' problem generally occur in pairs related by inversion, an odd number of solutions is possible in some cases, e.g., the single solution for PPP, or when one or three of the given circles are themselves solutions. (An example of the latter is given in the on Descartes' theorem.) However, there are no Apollonius problems with seven solutions. Alternative solutions based on geometry of circles and spheres have been developed and used for higher dimensions.cite journal | author = Knight RD | year = 2005 | title = The Apollonius contact problem and Lie contact geometry | journal = Journal of Geometry | volume = 83 | pages = 137&ndash;152 | doi = 10.1007/s00022-005-0009-x]

Mutually tangent given circles: Soddy's circles and Descartes' theorem

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 the inscribed and circumscribed circles, and are called "Soddy's circles". [ cite journal | author = Eppstein D | year = 2001 | title = Tangent Spheres and Triangle Centers | journal = The American Mathematical Monthly | volume = 108 | pages = 63&ndash;66 | doi = 10.2307/2695679] This special case of Apollonius' problem is also known as the four coins problem. [cite journal | author = Oldknow A | year = 1996 | title = The Euler-Gergonne-Soddy Triangle of a Triangle | journal = The American Mathematical Monthly | volume = 103 | pages = 319&ndash;329 | doi = 10.2307/2975188 cite web|authorlink= Eric W. Weisstein|author= Weisstein, EW| title = Four Coins Problem | url = http://mathworld.wolfram.com/FourCoinsProblem.html | publisher = MathWorld | accessdate = 2008-10-06 ] The three given circles of this Apollonius problem form a Steiner 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 as Descartes' theorem. In a 1643 letter to Princess Elizabeth of Bohemia, [Descartes R, "Œuvres de Descartes, Correspondance IV", (C. Adam and P. Tannery, Eds.), Paris: Leopold Cert 1901. fr icon] René Descartes showed that

:$left\left( k_\left\{1\right\}+k_\left\{2\right\}+k_\left\{3\right\}+k_\left\{s\right\} ight\right)^\left\{2\right\} = 2, left\left( k_\left\{1\right\}^\left\{2\right\} + k_\left\{2\right\}^\left\{2\right\} + k_\left\{3\right\}^\left\{2\right\} + k_\left\{s\right\}^\left\{2\right\} ight\right)$

where "k""s" = 1/"r""s" and "r""s" are the curvature and radius of the solution circle, respectively, and similarly for the curvatures "k"1, "k"2 and "k"3 and radii "r"1, "r"2 and "r"3 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.

Descartes' theorem was rediscovered independently in 1826 by Jakob Steiner,cite journal | author = Steiner J | year = 1826 | title = Einige geometrische Betrachtungen | journal = Journal für die reine und angewandte Mathematik | volume = 1 | pages = 161&ndash;184, 252&ndash;288 | url=http://www.digizeitschriften.de/no_cache/home/jkdigitools/loader/?tx_jkDigiTools_pi1%5BIDDOC%5D=512237] in 1842 by Philip Beecroft,cite journal | author = Beecroft H | year = 1842 | title = Properties of Circles in Mutual Contact | journal = Lady’s and Gentleman’s Diary | volume = 139 | pages = 91&ndash;96
cite journal | author = Beecroft H | year = 1846 | title = Unknown title | journal = Lady’s and Gentleman’s Diary | pages = 51 ( [http://www.pballew.net/soddy.html MathWords online article] )] and again in 1936 by Frederick Soddy.cite journal | author = Soddy F | date = 20 June 1936 | title = The Kiss Precise | journal = Nature | volume = 137 | pages = 1021 | doi = 10.1038/1371021a0] Soddy published his findings in the scientific journal "Nature" 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 curvature "k" of the circle.

::For pairs of lips to kiss maybe::Involves no trigonometry.::'Tis not so when four circles kiss::Each one the other three.::To bring this off the four must be::As three in one or one in three.::If one in three, beyond a doubt::Each gets three kisses from without.::If three in one, then is that one::Thrice kissed internally.

::Four circles to the kissing come.::The smaller are the benter.::The bend is just the inverse of::The distance from the center.::Though their intrigue left Euclid dumb::There's now no need for rule of thumb.::Since zero bend's a dead straight line::And concave bends have minus sign,::The sum of the squares of all four bends::Is half the square of their sum.

Sundry extensions of Descartes' theorem have been derived by Daniel Pedoe.cite journal | author = Pedoe D | year = 1967 | title = On a theorem in geometry | journal = Amer. Math. Monthly | volume = 74| pages = 627&ndash;640 | doi = 10.2307/2314247]

Generalizations

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; the ordinary Apollonius' problem corresponds to a special case in which the crossing angle is zero for all three given circles. Another generalization is the dual of the first extension, namely, to construct circles with three specified tangential distances from the three given circles.

Apollonius' problem can be extended from the plane to the sphere and other quadratic surfaces. For the sphere, the problem is to construct all the circles (the boundaries of spherical caps) that are tangent to three given circles on the sphere.cite book | author = Carnot L | year = 1803 | title = Géométrie de position | publisher = Unknown publisher | location = Paris | pages = p. 415, &sect;356] cite journal | author = Vannson | year = 1855 | title = Contact des cercles sur la sphère, par la geométrie | journal = Nouvelles Annales de Mathématiques | volume = XIV | pages = 55&ndash;71 fr icon] This spherical problem can be rendered into a corresponding planar problem using stereographic 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 by Charles Dupin.

By solving Apollonius' problem repeatedly to find the inscribed circle, the interstices between mutually tangential circles can be filled arbitrarily finely, forming an Apollonian gasket, also known as a "Leibniz packing" or an "Apollonian packing". [cite journal | author = Kasner E, Supnick F | year = 1943 | title = The Apollonian packing of circles | journal = Proc. Natl. Acad. Sci. USA | volume = 29 | pages = 378&ndash;384 | doi = 10.1073/pnas.29.11.378] This gasket is a fractal, being self-similar and having a dimension "d" that is not known exactly but is roughly 1.3,cite journal | author = Boyd DW | year = 1973 | title = Improved Bounds for the Disk Packing Constants | journal = Aeq. Math. | volume = 9 | pages = 99&ndash;106 | doi = 10.1007/BF01838194
cite journal | author = Boyd DW | year = 1973 | title = The Residual Set Dimension of the Apollonian Packing | journal = Mathematika | volume = 20 | pages = 170&ndash;174
cite journal|last=McMullen|first= Curtis T|title= Hausdorff dimension and conformal dynamics III: Computation of dimension|url=http://abel.math.harvard.edu/~ctm/papers/home/text/papers/dimIII/dimIII.pdf|journal=American Journal of Mathematics|volume=120|year=1998|pages=691&ndash;721|format=PDF|doi=10.1353/ajm.1998.0031] which is higher than that of a regular (or rectifiable) curve ("d" = 1) but less than that of a plane ("d" = 2). The Apollonian gasket was first described by Gottfried Leibniz in the 17th century, and is a curved precursor of the 20th-century Sierpiński triangle. [cite book | author = Mandelbrot B | year = 1983 | title = The Fractal Geometry of Nature | publisher = W. H. Freeman | location = New York | isbn = 978-0716711865 | pages = p. 170
cite book | author = Aste T, Weaire D | year = 2008 | title = In Pursuit of Perfect Packing | edition = 2nd edition | publisher = Taylor and Francis | location = New York | isbn = 978-1420068177 | pages = pp. 131&ndash;138
] The Apollonian gasket also has deep connections to other fields of mathematics; for example, it is the limit set of Kleinian groups. [cite book | author = Mumford D, Series C, Wright D | year = 2002 | title = Indra's Pearls: The Vision of Felix Klein | publisher = Cambridge University Press | location = Cambridge | isbn = 0-521-35253-3 | pages = pp. 196&ndash;223]

The configuration of a circle tangent to "four" circles in the plane has special properties, which have been elucidated by Larmor (1891)cite journal | author = Larmor A | year = 1891 | title = Contacts of Systems of Circles | journal = Proc. London Math. Soc. | volume = 23 | pages = 136&ndash;157 | doi = 10.1112/plms/s1-23.1.135] and Lachlan (1893).cite book | author = Lachlan R | year = 1893 | title = An elementary treatise on modern pure geometry | publisher = Macmillan | location = London | id = ASIN|B0008CQ720 | pages = §383&ndash;396, pp. 244&ndash;251] Such a configuration is also the basis for Casey's theorem, itself a generalization of Ptolemy's theorem.

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. For example, the given and solution spheres can be resized so that one given sphere is shrunk to point while maintaining tangency. 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 by Pierre de Fermat, [de Fermat P, "Varia opera mathematica", p. 74, Tolos, 1679.] and many alternative solution methods have been developed over the centuries.cite journal | author = Euler L | year = 1810 | title = Solutio facilis problematis, quo quaeritur sphaera, quae datas quatuor sphaeras utcunque dispositas contingat | journal = Memoires de l'academie des sciences de St.-Petersbourg | volume = 2 | pages = 17&ndash;28 | url = http://www.math.dartmouth.edu/~euler/docs/originals/E733.pdf la icon Reprinted in Euler's "Opera Omnia", series 1, volume 26, pp. 334&ndash;343.
cite book | author = Carnot L | year = 1803 | title = Géométrie de position | publisher = Imprimerie de Crapelet, chez J. B. M. Duprat | location = Paris | pages = p. 357, &sect;416 fr icon
cite journal | author = Hachette JNP | date = September 1808 | title = Sur le contact des sphères; sur la sphère tangente à quatre sphères données; sur le cercle tangent à trois cercles donnés | journal = Correspondance sur l'École Polytechnique | volume = 1 | issue = 2 | pages = pp. 27&ndash;28 fr icon
cite journal | author = Français J | date = January 1810 | title = De la sphère tangente à quatre sphères données | journal = Correspondance sur l'École Impériale Polytechnique | volume = 2 | issue = 2 | pages = pp. 63&ndash;66 fr icon
cite journal | author = Français J | date = January 1813 | title = Solution analytique du problème de la sphère tangente à quatre sphères données | journal = Correspondance sur l'École Impériale Polytechnique | volume = 2 | issue = 5 | pages = pp. 409&ndash;410 fr icon
cite journal | author = Dupin C | date = January 1813 | title = Mémoire sur les sphères | journal = Correspondance sur l'École Impériale Polytechnique | volume = 2 | issue = 5 | pages = p. 423 fr icon
cite book | author = Reye T | year = 1879 | title = Synthetische Geometrie der Kugeln | publisher = B. G. Teubner | location = Leipzig | url = http://www.gutenberg.org/files/17153/17153-pdf.pdf de icon
cite journal | author = Serret JA | year = 1848 | title = De la sphère tangente à quatre sphères donnèes | journal = Journal für die reine und angewandte Mathematik | volume = 37 | pages = 51&ndash;57 | url = http://www.digizeitschriften.de/index.php?id=loader&tx_jkDigiTools_pi1%5BIDDOC%5D=510729 fr icon
cite journal | author = Coaklay GW | date = 1859&ndash;1860 | title = Analytical Solutions of the Ten Problems in the Tangencies of Circles; and also of the Fifteen Problems in the Tangencies of Spheres | journal = The Mathematical Monthly | volume = 2 | pages = 116&ndash;126
cite journal | author = Alvord B | year = 1882 | title = The intersection of circles and intersection of spheres | journal = American Journal of Mathematics | volume = 5 | pages = 25&ndash;44, with four pages of Figures | doi = 10.2307/2369532
]

Apollonius' problem can even be extended to "d" dimensions, to construct the hyperspheres tangent to a given set of "d" + 1 hyperspheres. Following the publication of Frederick Soddy's re-derivation of the Descartes theorem in 1936, several people solved (independently) the mutually tangent case corresponding to Soddy's circles in "d" dimensions.cite journal | author = Gossett T | year = 1937 | title = The Kiss Precise | journal = Nature | volume = 139 | pages = 62 | doi = 10.1038/139062a0]

Applications

The principal application of Apollonius' problem and its generalizations to higher dimensions is trilateration, which seeks to determine a position from the "differences" in distances to at least three points. For example, a ship may seek to determine its position from the differences in arrival times of signals from three synchronized transmitters; conversely, the location of a homing beacon may be determined from the difference in arrival times of its signals at three receiving stations. Trilateration is equivalent to Apollonius' problem, as formulated by Isaac Newton: to determine a point in a plane from its distances to three known points, or to determine a point in space from its distances to four known points. Solutions to Apollonius' problem were used for trilateration in World War I to determine the positions of artillery pieces from the time required to hear the sound of the gun firing at three different positions. Trilateration is a key component of modern navigational systems such as GPScite journal | author = Hoshen J | year = 1996 | title = The GPS Equations and the Problem of Apollonius | journal = IEEE Transactions on Aerospace and Electronic Systems | volume = 32 | pages = 1116&ndash;1124 | doi = 10.1109/7.532270] and the earlier LORAN and Decca Navigator System.cite journal | author = Schmidt, RO | year = 1972 | title = A new approach to geometry of range difference location | journal = IEEE Transactions on Aerospace and Electronic Systems | volume = AES-8 | pages = 821&ndash;835 | doi = 10.1109/TAES.1972.309614] Trilateration is also used to determine the position of calling animals (such as birds and whales), although Apollonius' problem does not pertain if the speed of sound varies with direction (i.e., the transmission medium not isotropic).Principia", Isaac Newton used his solution of Apollonius' problem to construct an orbit in celestial mechanics from the center of attraction and observations of tangent lines to the orbit corresponding to instantaneous velocity. The special case of the problem of Apollonius when all three circles are tangent is used in the Hardy–Littlewood circle method of analytic number theory to construct Hans Rademacher's contour for complex integration, given by the boundaries of an infinite set of Ford circles each of which touches several others. [cite book | author = Apostol TM | title = Modular functions and Dirichlet series in number theory | publisher = Springer-Verlag | location = New York | edition = 2nd ed. | isbn = 978-0-387-97127-8 | year = 1990] Finally, Apollonius' problem has been applied to some types of packing problems, which arise in disparate fields such as the error-correcting codes used on DVDs and the design of pharmaceuticals that bind in a particular enzyme of a pathogenic bacterium. [cite journal | author = Lewis RH, Bridgett S | year = 2003 | title = Conic Tangency Equations and Apollonius Problems in Biochemistry and Pharmacology | journal = Mathematics and Computers in Simulation | volume = 61 | pages = 101&ndash;114 | doi = 10.1016/S0378-4754(02)00122-2]

References

Further reading

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* Trans., introd., and notes by Paul Ver Eecke. fr icon

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External links

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* The eight solutions step-by-step.

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