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 straightedge 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 using Cartesian coordinates. Letting (x_{1},y_{1}), (x_{2},y_{2}), and (x_{3},y_{3}) denote the centers of the three given circles, with radii r_{1}, r_{2}, and r_{3} respectively, the center of the unknown circle (x,y) satisfies the three equations 

_{} 

The eight possible choices of signs in the expressions (r ± r_{i}), 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 seconddegree 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 ± r_{i}) 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 + r_{i}), 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 accomplish 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 reinvert the plane to locate the tangent circle in the original plane. 

All of these geometric operations can be carried out using just straightedge 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 analogous 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 

_{} 

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 

_{} 

where s_{1} = a + b + c, s_{2} = ab + ac + bc, and s_{3} = 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 

_{} 

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

_{} 

The only positive real root of this quadratic is 

_{} 

This gives the tangency solution shown in the figure below: 


In general, the 8th degree polynomial always factors into (at least) the form 256g(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 

_{} 

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

_{} 

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 

_{} 

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

_{} 

where "Area" is the area of the triangle. 
