Apollonius' problem is the construction of a circle tangent to each of three given objects, each of which may be a point, line, or circle. There are a total of ten cases. The two easiest involve three points or three lines, and the hardest involves three circles. Euclid solved the two easiest cases in his Elements, and the others (with the exception of the three circle problem), appeared in the Tangencies of Apollonius which was, however, lost. The general problem is, in principle, solvable by straightedge and compass alone.
The three-circle problem was solved by Viète (Boyer 1968), and the solutions are called Apollonius circles. In general position, there are up to eight solutions. The simplest solution is obtained by solving the three simultaneous quadratic equations
 |
(1)
|
 |
(2)
|
 |
(3)
|
in the three unknowns 
, 
, 
for the eight triplets of signs (Courant and Robbins 1996).
Expanding the equations gives
 |
(4)
|
for 
,
2, 3. Since the first term is the same for each equation, taking 
and 
gives
 |
(5)
|
 |
(6)
|
where
 |  |  |
(7)
|
 |  |  |
(8)
|
 |  |  |
(9)
|
 |  |  |
(10)
|
and similarly for 
,
,
and 
(where the 2 subscripts are replaced by 3s). Solving these two simultaneous linear
equations gives
 |  |  |
(11)
|
 |  |  |
(12)
|
which can then be plugged back into the quadratic equation (1) and solved using the quadratic formula.
An explicit radius formula can be given when the three circle centers 
, 
, and 
are not collinear. Write 
, 
, 
, and 
, where 
is the area of 
. Suppose the given radii 
, 
, and 
satisfy 
, 
, and 
. Choose signed radii 
, 
, and 
, with a positive sign for external tangency and a negative
sign for internal tangency, and put
![K=+/-sqrt([a^2-(v-w)^2][b^2-(w-u)^2][c^2-(u-v)^2])](/images/equations/ApolloniusProblem/NumberedEquation7.svg) |
(13)
|
and
![D=S^2-sum_(cyc)[a^2u^2-(-a^2+b^2+c^2)vw].](/images/equations/ApolloniusProblem/NumberedEquation8.svg) |
(14)
|
For 
,
the solution radius is
![rho=(2SK-sum_(cyc)[(a^2u+(v+w)vw)(-a^2+b^2+c^2)-2a^2u^3])/(4D),](/images/equations/ApolloniusProblem/NumberedEquation9.svg) |
(15)
|
where the sums are over simultaneous cyclic permutations of 
and 
(C. E. Lozada, pers. comm., Sep. 7, 2023). Retain the values with
.
The exceptional case 
can still have finite solutions and must be handled separately
using the original tangency equations.
Perhaps the most elegant solution is due to Gergonne. It proceeds by locating the six homothetic centers (three internal and three external) of the three given circles. These lie three by three on four lines (illustrated above). Determine the inversion poles of one of these with respect to each of the three circles and connect the inversion poles with the radical center of the circles. If the connectors meet, then the three pairs of intersections are the points of tangency of two of the eight circles (Petersen 1879, Johnson 1929, Dörrie 1965). To determine which two of the eight Apollonius circles are produced by the three pairs, simply take the two which intersect the original three circles only in a single point of tangency. The procedure, when repeated, gives the other three pairs of circles.
If the three circles are pairwise externally tangent, the two solutions are the Soddy circles, allowing a common tangent line as a limiting solution (Eppstein 2001).
Larmor (1891) and Lachlan (1893, pp. 244-251) consider the problem of four circles having a common tangent circle.
