Apollonius' Tangency Problem

In Book IV of The Elements, Euclid shows how to construct the circle
that passes through three given points, and also how to construct
a circle tangent to three given straight lines.  Apollonius of Perga
(born circa 261 BC) subsequently generalized this by showing how to 
find a circle tangent to three objects in the plane, where the objects
can be any combination of points, lines, and/or circles.  The most
general and difficult case is obviously the case of three circles,
which was covered in Book II of Apollonius' "On Tangencies".
Unfortunately this entire work is lost, but we have a description
of its contents from Pappus.

In accord with the Greek preference for rigorous Euclidean methods of
construction for plane problems, the original problem of Apollonius 
was to construct a circle tangent to three given circles using only 
straight-edge and compass.  This is equivalent to requiring an 
algebraic solution involving nothing higher than quadratic equations.
By direct algebraic methods (though not very elegant) we can easily
find a quadratic whose root is the radius r of the unknown circle
usiing cartesian coordinates.  Letting (x1,y1), (x2,y2), and (x3,y3) 
denote the centers of the three given circles, with radii r1, r2, 
and r3 respectively, the center of the unknown circle (x,y) satisfies 
the three equations

     (x - xi)^2 + (y - yi)^2 = (r +- ri)^2    i = 1,2,3

The eight possible choices of signs in the expressions (r+-ri), 
i=1,2,3 correspond to the eight possible types of solutions for a 
given set of circles.  (The type of solution is classified according 
to which of the three given circles are inside and outside the 
unknown circle.)  Taking the differences between pairs of these 
equations, the second-degree terms in x,y, and r cancel, and we are 
left with two linear equations in the three unknowns x,y,r.  Solving 
these equations simultaneously for x and y and in terms of r, and 
substituting these expressions into one of the above formulas gives 
a quadratic in the unknown radius r.

If the choice of signs in the expressions (r+-ri) corresponds to an 
impossible type of solution for the given circles, neither root of 
the quadratic in r will be a positive real number.  On the other 
hand, if both of the roots are positive real numbers, then there 
are TWO circles of that particular type.  For example, it's possible
for two distinct circles to be externally tangent to three given
circles, and these two solutions would be the conjugate roots of
a single quadratic based on the system of equations with (r+ri), 
i=1,2,3.

Another way of solving Apollonius' problem is via inversive geometry.
We can invert the entire plane relative to a circle of radius R 
centered at the origin by allowing each point to remain along the
same direction from the origin, but changing the magnitude m of the
vector to R^2/m, where R is the radius of the inversion circle.
It's easy to see that this transformation maps circles to circles 
(regarding straight lines as circles of infinite radius), and any 
circle in the original plane that passes through the origin maps to 
a straight line in the transformed plane.  Also, tangency of curves 
is preserved under inversion.  

With these facts in mind, we can approach Apollonius' problem by 
first increasing the radius of each of the three given circles 
by a fixed amount such that two of the circles are just touching 
each other.  Obviously the center of the unknown circle for this 
configuration is the same as for the original configuration, so 
it suffices to solve this problem.  The advantage of increasing
all the radii until two of the given circle are touching is that 
we can now invert the plane about a circle centered on this point 
of tangency, and in the inverted plane those tangent circles will
be parallel straight lines.  

For convenience we can choose the radius R of our inversion circle 
so that it leaves the third given circle invariant.  To do this, 
our inversion circle must strike the third circle at right angles, 
which we can accommplish by drawing a circle on the segment from 
the center of inversion to the center of the third circle, and 
then noting the intersection of this curve with the third circle.
Once we have done this we need only find the circle tangent to the 
two parallel lines and to the third given circle, which is trivial.
Then we re-invert the plane to locate the tangent circle in the 
original plane.

All of these geometric operations can be carried out using just
straight-edge and compass, so this is equivalent to a solution by 
means of quadratic equations alone, just like the previous, purely
algebraic, method.

A third method is a hybrid of geometric and algebraic ideas.  From the
geometric standpoint we note that the area of the triangle formed by 
the centers of the three given circles equals the sum of the areas of 
the three triangles whose bases are edges of the original triangle, 
and whose remaining vertices are at the center of the unknown circle.
Using Heron's formula we can express all these areas in terms of the 
radii of the three given circles and the distances separating their 
centers, along with the radius r of the unknown circle.  This gives 
an equation in a single unknown, so we can theoretically solve for 
the unknown radius.  (This approach has been explored by Gene Dearing,
who has also extended it to cover the analagous problem for tangent
spheres in 3D space based on the volumes of tetrahedra.)

Unfortunately, since Heron's formula involves a square root, the 
equation that we produce with this method is of the form
              _      ____      ____      ____
             /K  =  /a(r)  +  /b(r)  +  /c(r)                      (1)

where K is the area of the main triangle and a(r), b(r), c(r) are 
each polynomials of degree 2 giving the areas of the subsidiary 
triangles.  As shown in Polynomials For Sums of Square Roots, an
equation of this form, when cleared of radicals, leads to the 
polynomial

          [(K - s1)^2 - 4 s2]^2  -  64 K s3  =  0                  (2)

where s1=a+b+c, s2=ab+ac+bc, and s3=abc.  Since each of a,b,c is a 
polynomial in the unknown quantity r of degree 2, the resulting 
polynomial is of degree 8, and it is extremely laborious to actually
generate this polynomial, let alone solve it.  For example, consider
the case of three circles whose centers are separated by distances of
32, 26, and 29 units, and whose radii are 4, 6, and 7 units (opposite
the edges, respectively) as shown below.

            

Applying equation (1) and clearing the radicals using equation (2), we 
arrive at the polynomial in the unknown radius r of the tangent circle

 1025797908486400 r^8 + 45688891412326400 r^7 - 2079538385255385600 r^6

   - 90879753802632739840 r^5 + 1684702930564817281024 r^4

    + 59130359737934771257344 r^3 - 733653834154604643483648 r^2

     - 12647913606051885385777152 r + 142656922868498947664510976  =  0

Fortunately, this polynomial can be factored, and the relevant root is 
a root of the quadratic

          2001755 r^2 + 23480730 r - 510609906  =  0

The only positive real root of this quadratic is
                                 _____
              11740365 + 186093 /33495
      r  =    ------------------------  =  11.14906897611679...
                      2001755

This gives the tangency solution shown in the figure below:

            

In general, the 8th degree polynomial ALWAYS factors into (at least)
the form  256 g(r) f(r) = 0,  where g(r) is of degree 6 and f(r) is
of degree 2, and the relevant root(s) are the roots of f(r).  These
represent the one or two possible solutions of a given "type".  In 
general the quadratic polynomial for r is

                c2 r^2  +  c1 r  +  c0  =  0

where the coefficients are given in terms of the parameters A,B,C,
and a,b,c by

     c2  =  4 [ (a-b)(a-c)A^2 + (b-a)(b-c)B^2 + (c-a)(c-b)C^2 ]

                 - (A+B+C)(-A+B+C)(A-B+C)(A+B-C)


 c1 =  2A^2 [(-a+b+c)(-a^2 + b^2 + c^2) - (-a^3 + b^3 + c^3) - a(-A^2 + B^2 + C^2)]

     + 2B^2 [( a-b+c)( a^2 - b^2 + c^2) - ( a^3 - b^3 + c^3) - b( A^2 - B^2 + C^2)]

     + 2C^2 [( a+b-c)( a^2 + b^2 - c^2) - ( a^3 + b^3 - c^3) - c( A^2 + B^2 - C^2)]


 c0 = (ABC)^2 - (A^2 B^2 + A^2 C^2 + B^2 C^2)(a^2 + b^2 + c^2)

              - (a^2 b^2 + a^2 c^2 + b^2 c^2)(A^2 + B^2 + C^2)

              + ((ABc)^2 + (AbC)^2 + (aBC)^2) + 2((abC)^2 + (aBc)^2 + (Abc)^2)

              + (A^4 a^2 + B^4 b^2 + C^4 c^2) + (a^4 A^2 + b^4 B^2 + c^4 C^2)

In the special case when a=b=c=0, all three of the given circles 
are points, and we are finding the radius of the circle that
circumscribes the triangle made by those three points, i.e., a
triangle with edges lengths A,B,C.  In this case the coefficients
reduce to

    c2  = -(A+B+C)(-A+B+C)(A-B+C)(A+B-C)

    c1 =  0

    c0 = (ABC)^2 

This gives the well-known formula for the radius of the circle
which circumscribes a triangle with edges lengths A,B,C

                           ABC                            ABC
    r   =  -----------------------------------    =     -------
           sqrt[(A+B+C)(-A+B+C)(A-B+C)(A+B-C)]          4[Area]

where "Area" is the area of the triangle.