Problem of Apollonius

Figure 3: Two given circles (black) and a circle tangent to both (pink). The center-to-center distances d₁ and d₂ equal r₁ + r_s and r₂ + r_s, respectively, so their difference is independent of r_s.

The solution of Adriaan van Roomen (1596) is based on the intersection of two hyperbolas.[13][14] 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 3). 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 (1687) refined van Roomen's solution, so that the solution-circle centers were located at the intersections of a line with a circle.[15] Newton formulates Apollonius' problem as a problem 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.[31] These four points correspond to the center of the solution circle (Z) and the centers of the three given circles (A, B and C).

The set of points with a constant ratio of distances d₁/d₂ to two fixed points is a circle.

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.

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.[32] Viète solved all ten of these cases using only compass and straightedge constructions, and used the solutions of simpler cases to solve the more complex cases.[1][20]

Figure 4: Tangency between circles is preserved if their radii are changed by equal amounts. A pink solution circle must shrink or swell with an internally tangent circle (black circle on the right), while externally tangent circles (two black circles on left) do the opposite.

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 Δr, the radius of its internally tangent given circles must be likewise changed by Δr, whereas the radius of its externally tangent given circles must be changed by −Δr. Thus, as the solution circle swells, the internally tangent given circles must swell in tandem, whereas the externally tangent given circles must shrink, to maintain their tangencies.

Viète used this approach to shrink one of the given circles to a point, thus reducing the problem to a simpler, already solved case. He first solved the CLL case (a circle and two lines) by shrinking the circle into a point, rendering it an 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.

Apollonius' problem can be framed as a system of three equations for the center and radius of the solution circle.[33] Since the three given circles and any solution circle must lie in the same plane, their positions can be specified in terms of the (x, y) coordinates of their centers. For example, the center positions of the three given circles may be written as (x₁, y₁), (x₂, y₂) and (x₃, y₃), whereas that of a solution circle can be written as (x_s, y_s). Similarly, the radii of the given circles and a solution circle can be written as r₁, r₂, r₃ 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:

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

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

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

The three numbers s₁, s₂ and s₃ 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_s² + y_s² on the left-hand side, and r_s² on the right-hand side. Subtracting one equation from another eliminates these quadratic terms; the remaining linear terms may be re-arranged to yield formulae for the coordinates x_s and y_s

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

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

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 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 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, 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 does not correspond to a geometrically possible solution for Apollonius' problem, since a solution circle cannot have an imaginary radius; therefore, the number of solutions is reduced by two. Apollonius' problem cannot have seven solutions, although it may have any other number of solutions from zero to eight.[12][34]

The same algebraic equations can be derived in the context of Lie sphere geometry.[26] 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:

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

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

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

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

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

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

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

where the vertical bars sandwiching c₁ − c₂ represent the length of that difference vector, i.e., the Euclidean norm. This formula shows that if two quadric vectors X₁ and X₂ are orthogonal (perpendicular) to one another—that is, 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

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

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

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

Therefore, Apollonius' problem can be re-stated in Lie geometry as a problem of finding perpendicular vectors on the Lie quadric; specifically, the goal is to identify solution 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.

${\displaystyle \left(X_{\mathrm {sol} }\mid X_{\mathrm {sol} }\right)=\left(X_{\mathrm {sol} }\mid X_{1}\right)=\left(X_{\mathrm {sol} }\mid X_{2}\right)=\left(X_{\mathrm {sol} }\mid X_{3}\right)=0}$

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 to higher-dimensional spaces.[26][35]

Figure 5: Inversion in a circle. The point P' is the inverse of point P with respect to the circle.

A natural setting for problem of Apollonius is inversive geometry.[4][12] The basic strategy of inversive methods is to transform a given Apollonius problem into another Apollonius problem that is simpler to solve; the solutions to the original problem are found from the solutions of the transformed problem by undoing the transformation. Candidate transformations must change one Apollonius problem into another; therefore, they must transform the given points, circles and lines to other points, circles and lines, and no other shapes. Circle inversion has this property and allows the center and radius of the inversion circle to be chosen judiciously. Other candidates include 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

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

Thus, if P lies outside the circle, then P' lies within, and vice versa. When P is the same as O, 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 an 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.[36]

Figure 6: A conjugate pair of solutions to Apollonius's problem (pink circles), with given circles in black.

Solutions to Apollonius's problem generally occur in pairs; for each solution circle, there is a conjugate solution circle (Figure 6).[1] One solution circle excludes the given circles that are enclosed by its conjugate solution, and vice versa. For example, in Figure 6, one solution circle (pink, upper left) encloses two given circles (black), but excludes a third; conversely, its conjugate solution (also pink, lower right) encloses that third given circle, but excludes the other two. The two conjugate solution circles are related 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.[4] For illustration, the orange circle in Figure 6 crosses the black given circles at right angles. Inversion in the radical circle leaves the given circles unchanged, but transforms the two conjugate pink solution circles into one another. Under the same inversion, the corresponding points of tangency of the two solution circles are transformed into one another; for illustration, in Figure 6, the two blue points lying on each green line are transformed into one another. Hence, the lines connecting these conjugate tangent points are invariant under the inversion; therefore, they must pass through the center of inversion, which is the radical center (green lines intersecting at the orange dot in Figure 6).

If two of the three given circles do not intersect, a center of inversion can be chosen so that those two given circles become concentric.[2][12] 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.

Figure 7: A solution circle (pink) in the first family lies between concentric given circles (black). Twice the solution radius r_s equals the difference r_outer − r_inner of the inner and outer radii, while twice its center distance d_s equals their sum.

Figure 8: A solution circle (pink) in the second family encloses the inner given circle (black). Twice the solution radius r_s equals the sum r_outer + r_inner of the inner and outer radii, while twice its center distance d_s equals their difference.

When two of the given circles are concentric, Apollonius's problem can be solved easily using a method of Gauss.[28] 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,

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

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

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

This formula represents four solutions, corresponding to the two choices of the sign of θ, and the two choices 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.

The usefulness of inversion can be increased significantly by resizing.[37][38] 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.

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.[37] In that case, Apollonius' problem degenerates to the CCP limiting case, 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.

In the second approach, the radii of the given circles are modified appropriately by an amount Δr so that two of them are tangential (touching).[38] Their point of tangency is chosen as the center 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.

Figure 9: The two tangent lines of the two tangent points of a given circle intersect on the radical axis R (red line) of the two solution circles (pink). The three points of intersection on R are the poles of the lines connecting the blue tangent points in each given circle (black).

Gergonne's approach is to consider the solution circles in pairs.[1] Let a pair of solution circles be denoted as C_A and C_B (the pink 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 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 6). 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 6).

To locate a second point on the lines L₁, L₂ and L₃, 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 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 9). 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 10).

Figure 10: The poles (red points) of the radical axis R in the three given circles (black) lie on the green lines connecting the tangent points. These lines may be constructed from the poles and the radical center (orange).

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 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

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

which implies 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.

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

Every quadratic equation in X, Y, and Z determines a unique conic, its vanishing locus. Conversely, every conic in the complex projective plane has an equation, and that equation is unique up to an overall scaling factor (because rescaling an equation does not change its vanishing locus). Therefore, the set of all conics may be parametrized by five-dimensional projective space P⁵, where the correspondence is

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

A circle in the complex projective plane is defined to be a conic that passes through the two points O₊ = [1 : i : 0] and O₋ = [1 : −i : 0], where i denotes a square root of −1. The points O₊ and O₋ are called the circular points. The projective variety of all circles is the subvariety of P⁵ consisting of those points which correspond to conics passing through the circular points. Substituting the circular points into the equation for a generic conic yields the two equations

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

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

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

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

Therefore, the variety of all circles is a three-dimensional linear subspace of P⁵. After rescaling and completing the square, these equations also demonstrate that every conic passing through the circular points has an equation of the form

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

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

Let D be a fixed smooth circle. If C is any other circle, then, by the definition of a circle, C and D intersect at the circular points O₊ and O₋. Because C and D are conics, Bézout's theorem implies C and D intersect in four points total, when those points are counted with the proper intersection multiplicity. That is, there are four points of intersection O₊, O₋, P, and Q, but some of these points might collide. Appolonius' problem is concerned with the situation where P = Q, meaning that the intersection multiplicity at that point is 2; if P is also equal to a circular point, this should be interpreted as the intersection multiplicity being 3.

Let Z_D be the variety of circles tangent to D. This variety is a quadric cone in the P³ of all circles. To see this, consider the incidence correspondence

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

For a curve that is the vanishing locus of a single equation f = 0, the condition that the curve meets D at r with multiplicity m means that the Taylor series expansion of f|_D vanishes to order m at r; it is therefore m linear conditions on the coefficients of f. This shows that, for each r, the fiber of Φ over r is a P¹ cut out by two linear equations in the space of circles. Consequently, Φ is irreducible of dimension 2. Since it is possible to exhibit a circle that is tangent to D at only a single point, a generic element of Z_D must be tangent at only a single point. Therefore, the projection Φ → P² sending (r, C) to C is a birational morphism. It follows that the image of Φ, which is Z_D, is also irreducible and two dimensional.

To determine the shape of Z_D, fix two distinct circles C₀ and C_∞, not necessarily tangent to D. These two circles determine a pencil, meaning a line L in the P³ of circles. If the equations of C₀ and C_∞ are f and g, respectively, then the points on L correspond to the circles whose equations are Sf + Tg, where [S : T] is a point of P¹. The points where L meets Z_D are precisely the circles in the pencil that are tangent to D.

There are two possibilities for the number of points of intersections. One is that either f or g, say f, is the equation for D. In this case, L is a line through D. If C_∞ is tangent to D, then so is every circle in the pencil, and therefore L is contained in Z_D. The other possibility is that neither f nor g is the equation for D. In this case, the function (f / g)|_D is a quotient of quadratics, neither of which vanishes identically. Therefore, it vanishes at two points and has poles at two points. These are the points in C₀ ∩ D and C_∞ ∩ D, respectively, counted with multiplicity and with the circular points deducted. The rational function determines a morphism D → P¹ of degree two. The fiber over [S : T] ∈ P¹ is the set of points P for which f(P)T = g(P)S. These are precisely the points at which the circle whose equation is Tf − Sg meets D. The branch points of this morphism are the circles tangent to D. By the Riemann–Hurwitz formula, there are precisely two branch points, and therefore L meets Z_D in two points. Together, these two possibilities for the intersection of L and Z_D demonstrate that Z_D is a quadric cone. All such cones in P³ are the same up to a change of coordinates, so this completely determines the shape of Z_D.

To conclude the argument, let D₁, D₂, and D₃ be three circles. If the intersection Z_D1 ∩ Z_D2 ∩ Z_D3 is finite, then it has degree 2³ = 8, and therefore there are eight solutions to the problem of Apollonius, counted with multiplicity. To prove that the intersection is generically finite, consider the incidence correspondence

${\displaystyle \Psi =\{(D_{1},D_{2},D_{3},C)\in (\mathbf {P} ^{3})^{4}\colon C{\text{ is tangent to all }}D_{i}\}.}$

There is a morphism which projects Ψ onto its final factor of P³. The fiber over C is Z_C³. This has dimension 6, so Ψ has dimension 9. Because (P³)³ also has dimension 9, the generic fiber of the projection from Ψ to the first three factors cannot have positive dimension. This proves that generically, there are eight solutions counted with multiplicity. Since it is possible to exhibit a configuration where the eight solutions are distinct, the generic configuration must have all eight solutions distinct.

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).[32] 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 (the incircle and excircles of the triangle formed by the three lines).

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

Table 1: Ten Types of Apollonius' Problem

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

Figure 11: An Apollonius problem with no solutions. A solution circle (pink) must cross the dashed given circle (black) to touch both of the other given circles (also black).

The problem of counting the number of solutions to different types of Apollonius' problem belongs to the field of enumerative geometry.[12][41] The general number of solutions for each of the ten types of Apollonius' problem is given in Table 1 above. However, special arrangements of the given elements may change the number of solutions. For illustration, Apollonius' problem has no solution if one circle separates the two (Figure 11); to touch both the solid given circles, the solution circle would have to cross the dashed given circle; but that it cannot do, if it is to touch the dashed circle tangentially. Conversely, if three given circles are all tangent at the same point, 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,[42] although earlier work had been done by Stoll[43] and Study.[44] However, Muirhead's work was incomplete; it was extended in 1974[45] and a definitive enumeration, with 33 distinct cases, was published in 1983.[12] Although solutions to Apollonius' problem generally occur in pairs related 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 section on Descartes' theorem.) However, there are no Apollonius problems with seven solutions.[34][43] Alternative solutions based on the geometry of circles and spheres have been developed and used in higher dimensions.[26][35]

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

Figure 12: The two solutions (red) to Apollonius' problem with mutually tangent given circles (black), labeled by their curvatures.

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,[48] René Descartes showed that

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

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.[2][49]

Descartes' theorem was rediscovered independently in 1826 by Jakob Steiner,[50] in 1842 by Philip Beecroft,[2][49] and again in 1936 by Frederick Soddy.[51] 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.

Sundry extensions of Descartes' theorem have been derived by Daniel Pedoe.[52]