Problem of Apollonius

Figure 1: The eight solutions of Apollonius' problem. The three given circles are shown in black. The eight solution circles are shown as four pairs (blue, magenta, yellow and grey). In each pair, one solution circle encloses the given circles that are excluded by the other solution, and vice versa. For example, the larger blue solution encloses the two larger given circles, but excludes the smallest; the smaller blue solution does the reverse.

One of the most famous problems in Euclidean plane geometry, the problem of Apollonius is to construct a circle that is tangent to ("touches") three given circles. The three given circles can be of any size and at any distance from one another; as limiting cases, they may even be points (circles of zero radius) or lines (circles of infinite radius). Apollonius of Perga (ca. 262 BC – ca. 190 BC) posed and solved the problem and its limiting cases in his work Επαφαι ("Tangencies", Latin: De tactionibus, De contactibus); although this work was lost, a 4th-century report of his results by Pappus of Alexandria has survived, from which a reconstruction of Apollonius' solution was developed by François Viète.

In general, Apollonius' problem has eight solutions, which differ in how they include or exclude the given circles (Figure 1). These solution circles—sometimes called Apollonius circles, although the term is ambiguous—may be found by a great variety of geometric and algebraic methods, including transformations such as circle inversion. The first modern solution was developed in the 16th century by Adriaan van Roomen (writing in Latin as Adrianus Romanus), but many eminent mathematicians have developed alternative methods, including Issac Newton, Carl Friedrich Gauss, Lazare Carnot, Augustin Louis Cauchy, Jean-Victor Poncelet and Joseph Diaz Gergonne. If the three given circles touch each other, their radii are related to those of the solutions by a theorem of René Descartes. This was rediscovered three centuries later by Nobel laureate Frederick Soddy, who published it as a poem.

Apollonius' problem has been generalized in several ways. The three given circles may be chosen to lie on a quadric surface, such as a sphere, instead of the plane. Another generalization is to construct a circle that crosses 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.

Although Apollonius' problem has historically served as a test for novel methods in geometry, it also has found practical applications, such as in celestial mechanics, trilateration, and packing problems. One of the earliest fractals described in print was the Apollonian gasket proposed by Gottfried Leibniz, which is based on solving Apollonius' problem iteratively.

Statement of Appolonius' problem

The precise statement of Apollonius' problem is: Let C₁, C₂, and C₃ be three circles in the plane. They are allowed to be degenerate, meaning that they may be lines (considered as circles whose center is at infinity and whose radius is infinite) or points (considered as circles of radius zero). They are also allowed to cross each other and even to be identical (to lie on top of each other). Apollonius' problem asks, "How many circles are simultaneously tangent to C₁, C₂, and C₃?" (Two circles are tangent at a point if they intersect at that point and the angle formed by the two circles at that point is zero.)

Circle inversion and pairing of solutions

Figure 2: P ' is the inverse of P with respect to the circle.

If one of the given circles is already tangent to the other two, then it is a solution. When this does not happen, the solutions to Appolonius' problem come in pairs because any solution can be inverted to produce another solution. Inversion in a circle with center O and radius R is the following operation (Figure 2): Every point P is mapped into a new point P' such that O, P, and P' all lie on the same line, and the product of the distances of P and P' to the center O equal the radius R squared:

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

Thus if P lies outside the circle, then P' lies within, and vice versa. When P is the same as O, then the inversion is said to send P to infinity; if P is very close to O, then P' is very far, so when P equals O, P' ought to be at infinity in some sense. (In the field of complex analysis, "infinity" is made precise by working on the Riemann sphere; circle inversions are a type of Möbius transformation on the sphere.) Inverting again with respect to the same circle returns P to itself. Circle inversions are conformal maps, meaning that the angle between any two lines is the same as the angle between their inversion. This implies that a circle which does not pass through the center of the inversion circle is transformed into another circle. If a circle passes through the center of the inversion circle, then it is transformed into a straight line. These facts can be exploited to show that Appolonius' problem has an even number of solutions in this case, and below it will even be used to solve Apollonius' problem.

Figure 3: The pink and black circles (corresponding to the magenta circles in Figure 1) are a conjugate pair of solutions to Apollonius' problem. The given circles are drawn in red, green and blue. Inversion in the orange circle C_G (centered on the radical center G) leaves the given circles invariant, but transforms the pink and black solution circles into one another. Since their points of tangency map into one another, the three lines L₁, L₂, L₃ intersect at the radical center G.

Start with one solution to Appolonius' problem, the pink circle in figure 3. For any three distinct circles, such as the three given circles, it is always possible to find a fourth circle which intersects the given circles perpendicularly. Call this circle C_G. The inversion of the pink circle is another circle, the black circle. Inversion in C_G transforms each of the given circles into itself: A point on one of the given circles and inside C_G is sent to a point on the same given circle and outside C_G, and vice versa for points outside C_G. In particular, this is true for the points where the pink circle are tangent to the given circles: A₁, A₂, and A₃ are points on the first, second, and third given circles, so their inversions B₁, B₂, and B₃ also lie on the given circles. Because circle inversion is conformal, the angles made by the pink circle with the given circles at A₁, A₂, and A₃ must be the same as the angles made by the black circle with the given circles at B₁, B₂, and B₃. Therefore the black circle is also tangent to the three given circles, that is, it is also a solution. Inverting in C_G again returns the new solution to the old one, and this correspondence shows that solutions come in pairs.

Solutions in special cases

Two or three identical circles

When all three circles are identical, then Appolonius' problem is equivalent to asking for all the circles tangent to a given circle. There are an infinite number of such circles, one for every point in the plane: Given such a point P, find the distance from the point to the given circle, and subtract the radius of the given circle. If this number is negative, take its absolute value. Take the result to be the radius of a circle centered at P; this circle is tangent to the given circle. When P lies on the given circle, the circle at P has radius zero, so it is a point. When P is the center of the given circle, the circle centered at P is identical with the given circle. (A circle is assumed to be tangent to itself.)

When two of the circles are identical, Appolonius' problem again has infinitely many solutions. However, the centers of the solution circles all lie on a hyperbola. This fact was used by Adriaan van Roomen to solve the general case of Appolonius' problem, and it is discussed below.

For the rest of this article, the three circles are assumed to be distinct.

Mutually tangent given circles: Descartes' theorem

File:DescartesCircles.png

Figure 4: If the given three circles (shown in black) are mutually tangent, Apollonius' problem has only two other solutions (shown in red). The inner solution circle "gets three kisses from without"; the outer circle is "thrice kissed internally". Each circle is labeled with its curvature.

If the three given circles are mutually tangent, Apollonius' problem has five solutions, three of which are the given circles themselves; as before, circles are tangent to themselves. The other two solutions, shown in red in Figure 4, are the inscribed and circumscribed circles. Either of these latter two solutions, together with the three given circles, produces a set of four circles that are mutually tangent at six points. The radii of these circles are related by an equation known as Descartes' theorem. In a 1643 letter to Princess Elizabeth of Bohemia, René Descartes showed that

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

where k_s = 1/r_s and r_s are the curvature and radius of the solution circle, respectively, and similarly for the curvatures k₁, k₂, and k₃ and radii r₁, r₂, and r₃ 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, in 1842 by an amateur mathematician, Philip Beecroft, and again in 1936 by Nobel laureate Frederick Soddy. 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.

Limiting cases: points and lines

Main article: Limiting cases of Apollonius' problem

The general Apollonius problem is to find a circle tangent to three given circles. However, there are nine limiting cases (Table 1), in which at least one of the given circles shrinks to zero radius (a point) or swells to infinite radius (a straight line). In general, the first seven limiting cases have fewer solutions than the general problem. These ten types (the general problem and the nine limiting cases) are labeled with three letters, either C, L or P, to denote whether the given elements are a circle, line or point, respectively; for example, the type of Apollonius problem with a given circle, line and point is denoted as CLP. The problems of finding a circle tangent to three points (PPP) and to three lines (LLL) were solved first by Euclid.

Table 1: Ten Types of Apollonius' Problem

Index	Code	Given Elements	Number of solutions (in general)	Example (solution=pink; givens=red/green/blue)
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

Figure 5: An Apollonius problem with no solutions.

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 5); to touch both the red and blue circles, the solution circle would have to cross the green circle; but that it cannot do, if it is to touch the green 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; thus, such Apollonius problems have an infinite number of solutions. 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, although earlier work had been done by Stoll and Study. However, Muirhead's work was incomplete; it was extended in 1974 and a definitive enumeration was published in 1983. One interesting result is that there are no configurations of circles that give rise to seven solutions. Alternative solutions based on Lie geometry have been developed and used for higher dimensions.

Solution methods

Apollonius' problem has had a rich history and numerous solution methods have been developed over the centuries. The original geometrical method used by Apollonius of Perga has been lost, but reconstructions have been offered by François Viète and by T. L. Heath, based on the clues in the description by Pappus. After its re-introduction in the 16th century, Adriaan van Roomen solved for the centers of the solution circles as the intersection of two hyperbolae, a method that was refined by Isaac Newton in his Principia; Newton's solution was re-discovered independently by John Casey in 1881. Apollonius' problem has also been solved algebraically; this approach was pioneered by René Descartes and Princess Elisabeth of Bohemia, and subsequently refined by Leonhard Euler, Nicolas Fuss, Carl Friedrich Gauss, Lazare Carnot, Augustin Louis Cauchy and Jean-Victor Poncelet. Direct geometrical solutions have been published by Joseph Diaz Gergonne and by Poncelet, with alternative formulations being offered by Maurice Fouché and others. Methods using circle inversion were pioneered by Julius Petersen and are sometimes considered to be the most intuitive approach for lay-people; in some versions, the solution circle is transformed into a line, or confined between two lines or two concentric circles. These various approaches are outlined below.

Figure 6: The difference in center-to-center distances d₁ and d₂ between the solution circle (black) and two given circles (red and blue) does not depend on the radius r_s of the solution circle; it depends only on the difference of the given radii, r₂−r₁.

Intersecting hyperbolas

The first solution of Apollonius' problem in modern times was that of Adriaan van Roomen and is based on the intersection of two hyperbolas. Let the given circles be denoted as C₁, C₂ and C₃. 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 C₁ and C₂. 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, r₁ and r₂, respectively (Figure 6). The distance d₁ between the centers of the solution circle and C₁ is either r_s + r₁ or r_s − r₁, depending on whether these circles are chosen to be externally or internally tangent, respectively. Similarly, the distance d₂ between the centers of the solution circle and C₂ is either r_s + r₂ or r_s − r₂, again depending on their chosen tangency. Thus, the difference d₁ − d₂ 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 C₂ and C₃, where the internal or external tangency of the solution and C₂ 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 refined van Roomen's solution, so that the solution-circle centers were located at the intersections of a line with a circle. Newton phrases Apollonius' problem slightly differently, namely, 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, by Apollonius' definition of the circle, Z also lies on a known circle. Thus, the solutions to Apollonius' problem are the intersections of a line with a circle. Newton's solution was re-discovered independently by John Casey in 1881.

Straightedge and compass solutions

A prized property in classical Euclidean geometry was solution of a problem using compass and straightedge constructions. Many relatively basic constructions are not possible using only these tools, for example angle trisection and doubling the cube. In general, a geometrical construction can be done with compass and straightedge if and only if the ratios of its distances involve only square roots and not cube roots, higher roots, or transcendental numbers such as π. Van Roomen's solution uses the intersection of two conics, and therefore does not satisfy the ancient Greek requirement. The mathematician Regiomontanus doubted whether the problem could be solved under such restrictions.

Van Roomen's friend François Viète, who had originally urged van Roomen to work on the problem in the first place, noted noted that many problems which are impossible to solve using a ruler and straightedge are possible by intersecting conics, for instance, doubling the cube cannot be done using a straightedge and compass, but Menaechmus showed that it is possible to do using the intersections of two parabolas. Viète produced a compass and straightedge construction for all ten special cases of Apollonius' problem. His methods often shrink a given circle to a point to reduce the problem to a simpler special case (see below).

Algebraic solutions

Figure 7: The black circle is one solution of Apollonius' problem, being internally tangent to the green given circle C₂ and externally tangent to the red and blue circles, C₁ and C₃. The signs for this solution are "− + −"; the solution encloses C₂ and excludes C₁ and C₃.

Apollonius' problem can be framed as a system of three coupled quadratic equations in three variables x_s, y_s and r_s

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

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

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

Here, r₁, r₂, and r₃ represent the radii of the three given circles, whereas (x₁, y₁), (x₂, y₂), and (x₃, y₃) represent their center positions in Cartesian coordinates. The three signs s₁, s₂, and s₃ on the right-hand side may equal ±1, giving eight possible sets of equations (2 × 2 × 2 = 8), each one corresponding to one of the eight types of solution circles. If a particular sign (say, s₁) is positive, the solution circle will be internally tangent to the corresponding given circle (in this case, C₁); the solution circle encloses the given circle. Conversely, if a particular sign is negative, the solution is externally tangent to the corresponding given circle; in other words, the solution circle excludes the given circle (Figure 7).

Whatever the choice of signs, this system of three equations may be solved by the method of resultants. Multiplying out the three equations and canceling the common terms yields formulae for the coordinates x_s and y_s

$x_{s}=M+Nr_{s}$

$y_{s}=P+Qr_{s}$

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 in 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.

Algebraic solutions to Apollonius' problem were pioneered by René Descartes and Princess Elisabeth of Bohemia, although their solutions were rather complex. The equations were subsequently refined by Leonhard Euler, Nicolas Fuss, Carl Friedrich Gauss, Lazare Carnot, Augustin Louis Cauchy and Jean-Victor Poncelet.

The signs s₁, s₂, and s₃ 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 is a solution, so is −r_s; these represent the same circle, but with opposite signs s_i. Therefore, Apollonius' problem has at most eight independent solutions, the number predicted by Bézout's theorem. These eight types of solutions are depicted in Figure 1.

Lie sphere geometry

The same algebraic equations can be derived in the context of Lie sphere geometry. 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 is imagined as an "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 an unusual product similar to the dot product:

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

The Lie quadric is defined as those vectors whose product with themselves (their norm) is zero, (X|X) = 0. Let X₁ and X₂ be two vectors belonging to this quadric; the norm of their difference equals

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

The product distributes over addition and subtraction:

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

Since (X₁|X₁) = (X₂|X₂) = 0 (both belong to the Lie quadric) and since w₁ = w₂ = 1 for circles, the product of any two such vectors on the quadric equals

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

This formula shows that if two quadric vectors X₁ and X₂ are orthogonal (perpendicular) to one another—mathematically, if (X₁|X₂) = 0—then their corresponding circles are tangent. For if the two signs s₁ and s₂ 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

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

Conversely, if the two signs s₁ and s₂ 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

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

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 vectors X_sol that belong to the Lie quadric and are also orthogonal (perpendicular) to the vectors X₁, X₂, and X₃ corresponding to the given circles.

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

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 into higher dimensional spaces.

Inversive methods

The basic approach 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. One must transform circles and lines to circles and lines, and only so many motions of the plane meet this restriction. Candidates include the Euclidean plane isometries, which are built from translations, rotations, and reflections across straight lines. These isometries are not powerful enough to substantially simplify the problem. However, circle inversion makes it possible to turn an Apollonius problem into a special case which is easier to solve.

The application of inversion to Apollonius' problem was pioneered by Julius Petersen. Other inversive solutions are possible besides the common ones described below.

Figure 8: The solutions (pink circle) of the first family lie between the concentric given circles (red and blue). There are four solutions, depending on the sign of θ and whether the solution circle is internally or externally tangent to the non-concentric given circle (green).

Inversion to an annulus

If two of the three given circles are disjoint, 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 8), the solutions do not enclose the inner concentric circle, but rather revolve like ball bearings in the annulus. In the second family (Figure 9), the solution circles enclose the inner concentric circle. There are generally four solutions for each family, yielding eight possible solutions in all, consistent with the algebraic solution.

Figure 9: The solutions (pink) of the second family touch the inner and outer concentric circles, but enclose the inner circle. There are again four solutions.

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 8). 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 8), 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. Then by the law of cosines,

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

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

$\theta =\pm 2\ \mathrm {atan} \left({\sqrt {\frac {1-C_{\pm }}{1+C_{\pm }}}}\right).$

The other four solutions can be obtained by the same method, with the substitutions for r_s and d_s indicated in Figure 9.

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

Figure 10: The tangency of a set of circles is preserved if their radii are changed by the same amount. Two internally tangent circles must shrink or swell together to preserve their tangency, whereas two externally tangent circles must do the opposite: if one circle shrinks in radius by Δr, the other must swell by the same amount.

In some versions of the inversive approach, the given circles are resized to some special condition prior to the inversion. The solution circle will remain tangent under this resizing if it too expands or contracts appropriately. 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. The former must swell with the solution circle, while the latter must shrink, to maintain their tangency (Figure 10).

Thus, the initial Apollonius problem can be transformed into another problem that satisfies some constraint on the given circles. For example, any set of four Apollonius circles (the three given circles and a particular solution) can be resized so that the smallest given circle is shrunk to a point. Alternatively, two given circles can often be resized so that they are tangent to one another. Thirdly, intersecting given circles can be resized so that they become non-intersecting, so that the above 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.

Shrinking 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. In that case, Apollonius' problem degenerates to finding a solution circle tangent to the two remaining given circles that passes through the point P, one of the simpler limiting cases. 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. To obtain the remaining solutions, different combinations of shrinking and swelling the radii according to the signs will result in different given circles being shrunk to a point.

Resizing two given circles to tangency

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). 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 Δr produces a solution circle tangent to the original three circles.

Gergonne's solution

Figure 11: Gergonne noticed that the two tangent lines drawn from any of the given circles intersected on the radical axis R of the two solution circles. The three points of intersection are the poles of L_1–3 in their respective given circles C_1–3.

The solution to Apollonius' problem published by Joseph Diaz Gergonne in 1814 is widely considered to be the most elegant. 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 4), 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 L₁ 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 L₁ with the given circle C₁ (Figure 4). The remaining four tangent points would be located similarly, by finding lines L₂ and L₃ that contained A₂ and B₂, and A₃ and B₃, respectively. To construct a line such as L₁, 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 4).

Figure 12: Conversely, the respective poles P_1–3 (shown in orange) of R in the three given circles C_1–3 must lie on the lines L_1–3. Together with the radical center G, these poles define the lines L_1–3.

To locate a second point on the lines L_1–3, Gergonne noted a reciprocal relationship between the radical axis R of the solution circles, C_A and C_B, and the lines L_1–3. To understand this reciprocal relationship, consider the two tangent lines to the circle C₁ drawn at its tangent points A₁ and B₁ with the solution circles; the intersection of these tangent lines is the pole point of L₁ in C₁. 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 11). The relationship between pole points and their polar lines is reciprocal; if the pole of L₁ in C₁ lies on R, the pole of R in C₁ must conversely lie on L₁. Thus, if we can construct R, we can find its pole P₁ in C₁, giving the needed second point on L₁ (Figure 12).

Gergone found the radical axis R of the unknown solution circles as noted. 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 C₁ and C₂; 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

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

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 L₁ is defined by two points: the radical center G of the three given circles and the pole in C₁ of one of the four lines connecting the homothetic centers. Finding the same pole in C₂ and C₃ gives L₂ and L₃, 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.

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.

Figure 13: A symmetrical Apollonian gasket, also called the Leibniz packing, after its inventor Gottfried Leibniz.

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. 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. This gasket is a fractal, being self-similar and having a dimension d that is roughly 1.3,, 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. The Apollonian gasket also has deep connections to other fields of mathematics; for example, it is the limit set of Kleinian groups.

The configuration of a circle tangent to four circles in the plane has special properties, which have been elucidated by Larmor (1891) and Lachlan (1893). 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, and many alternative solution methods have been developed over the centuries.

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.

Applications

Apollonius' problem is a pure problem in Euclidean plane geometry, and Apollonius himself was famously indifferent to the practical applications of his work.

[This] subject is one of those which seems worthy of study for their own sake.

— Apollonius of Perga, from the preface of Book V of his Conics

Nonetheless, Apollonius' problem and its generalizations to higher dimensions have found some applications. Its principal application lies in trilateration, which is the problem of determining a receiver position from the differences in distances to at least three transmitters or conversely, from the differences in distances between a transmitter and at least three receivers. This problem is equivalent to Apollonius' problem, as expressed by Isaac Newton: to determine a point in a place (resp. in space) from its distances to three (resp. 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 GPS and the earlier LORAN and Decca Navigator System, although the connection to Apollonius' problem is not always recognized. 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., is not isotropic).

Apollonius' problem has other applications as well. Isaac Newton used his solution to treat some problems of celestial mechanics, particular Proposition 21 in his Principia. It is also used in the Hardy-Littlewood circle method of analytic number theory describe a certain complicated contour for complex integration, given by the boundaries of an infinite set of circles each of which touches several others. 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.

Further reading

• Boyd DW (1973). "The osculatory packing of a three-dimensional sphere". Canadian J. Math. 25: 303–322.

• Coolidge JL (1916). A Treatise on the Circle and the Sphere. Oxford: Clarendon Press. pp. 167–172. Calls Apollonius' problem "the most famous of all" geometry problems.

• Dörrie H (1965). "The Tangency Problem of Apollonius". 100 Great Problems of Elementary Mathematics: Their History and Solutions. New York: Dover. pp. pp. 154-160 (§32). {{cite book}}: |pages= has extra text (help)

• Eppstein D (2001). "Tangent Spheres and Triangle Centers". The American Mathematical Monthly. 108: 63–66. doi:10.2307/2695679.

• Gisch D, Ribando JM (2004). "Apollonius' Problem: A Study of Solutions and Their Connections". American Journal of Undergraduate Research. 3: 15–25.

• Oldknow A (1996). "The Euler-Gergonne-Soddy Triangle of a Triangle". The American Mathematical Monthly. 103: 319–329. doi:10.2307/2975188.

• Simon M (1906). Über die Entwicklung der Elementargeometrie im XIX. Jahrhundert. Berlin: Teubner. pp. pp. 97–105. {{cite book}}: |pages= has extra text (help)

• Wells D (1991). The Penguin Dictionary of Curious and Interesting Geometry. New York: Penguin Books. pp. pp. 3–5. ISBN 0-14-011813-6. {{cite book}}: |pages= has extra text (help)

External links

• "Ask Dr. Math solution". Mathforum. Retrieved 2008-05-05.
• Weisstein, Eric W. "Apollonius' problem". MathWorld.
• "Apollonius' Tangency Problem". MathPages. Retrieved 2008-05-05.
• "Apollonius' Problem". Cut The Knot. Retrieved 2008-05-05. {{cite web}}: Text "cite web" ignored (help)
• Kunkel, Paul. "Tangent Circles". Whistler Alley. Retrieved 2008-05-05.
• Austin, David (March 2006). "When kissing involves trigonometry". Feature Column at the American Mathematical Society website. Retrieved 2008-05-05.
• "Apollonius' Tangency Problem For Three Circles". gogeometry.com. Retrieved 2008-05-05. The eight solutions step-by-step.