Given three objects, each of which may be a point, line, or circle, draw a circle that
is tangent to each. 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.
There are eight total 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
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 mutually tangent, then the eight
solutions collapse to two, known as the Soddy circles.
Larmor (1891) and Lachlan (1893, pp. 244-251) consider the problem of four circles having a common tangent circle.
, 
, 
for the eight triplets of signs (Courant and Robbins 1996).
Expanding the equations gives
,
2, 3. Since the first term is the same for each equation, taking 
and 
gives
,
,
and 
(where the 2 subscripts are replaced by 3s). Solving these two simultaneous linear
equations gives
