Problem of Apollonius

Apollonius of Perge is dedicated to the geometric problem to construct a circle tangent to any three other circles, a non- preserved book ( About touches ).

Since one can also start from an infinitely small radius and an infinitely large radius at the output circuits, it can be considered not only by three circles but also of points and lines ( tangents ). In total there are ten possible combinations for the given pieces that are listed below.

Since the complete solution of the problems all design cases solves touches ( tangents ) of circles, points and lines, of course, the Berührkreise are at the triangle containing ( Ankreis, inscribed circle, radius).

• 3.1 Overview
• 3.2 Three straight
• 3.3 Two lines, a point
• 3.4 A straight line, two points
• 3.5 Three points

History

The two simplest cases ( three points and three lines ) have already been solved by Euclid, the others were included in the lost works of Apollonius. The set of Descartes deals with the case of three circles which touch each other in pairs. François Viète solved the problem as the first in modern times by means of a nonlinear system of three quadratic equations. A very elegant solution comes from Joseph Gergonne.

Solution methods

Intersection of hyperbolas

The solution method of Adrian van Roomen (1596 ) is based on the intersection of hyperbolas, but does not constitute construction with ruler and compass represents the centers of the given circles are with, referred to, their radii, . Wanted is a circle with center and radius. Then each must be the same or equal to the distance - depending on whether the contact is exclusive or inclusive. The difference of the distances between two given and the circle center points must thus each have a certain value, which depends only on the given radius. In other words: The searched center of the circle must be on a certain hyperbola whose foci coincide with the given centers. Average of two hyperbolas of this kind one finds the center of this circle.

Algebraic solution

The conditions for the distances of the desired circle center of the given centers lead to a system of equations of the following type for the three unknowns, and:

With exclusive touch the plus sign of, in enclosing touch the minus sign applies. If one subtracts example, the second equation from the first and the third from the second, and so one can express through by solving a system of linear equations with two unknowns. By substituting the result into one of the given equations one obtains a quadratic equation with the can be determined.

Special cases

Overview

For the four simplest cases can be specified using relatively simple ways solutions for the radii:

Three straight

For three intersecting straight lines ( not parallel or one above the other ), there are four solutions. Are the two lines are parallel, there are only two solutions for three parallels there is no solution and for parallel lines with distance 0 there are infinitely many solutions.

The three lines, with their intersections a triangle with the sides. So here's the rules for the inscribed circle and the excircles are used:

With the internal angles, the surface area and half the circumference s:

To obtain an expression that uses only the side lengths, the set of Heron can be used:

Are the corresponding formulas for the excircles

And accordingly for the other Ankreisen.

Two straight lines, a point

There are different cases:

• The lines are parallel: if the point is outside the limited range of the straight line is, there are no solutions. If it lies on a straight line, there is a solution. Is he in between two solutions; the diameter in each case is equal to the distance of the straight line.
• The lines are not parallel: Is the point of intersection of the straight line, there is no solution.
• If the point lies on a straight line, but is not the point of intersection, there are two solutions; the centers of the circles are the points of intersection of the two with the perpendicular to the bisector through.
• If the point on any of the straight line, there are two solutions; this is the generic case. Let the image from among the reflection to the associated bisector and perpendicular to the intersection of the through, and with one of the straight lines. Then have the contact points of the two circles with the distance from. Their centers are the respective intersections of the perpendicular to the contact points so identified with.

A straight line, two points

For two points and a straight line, there are two solutions, a solution with the below mentioned special cases, only one, and two spaced points of the straight line.

This case is substantially equivalent to the previous; the perpendicular bisector of the line connecting the two points corresponds to the bisector. Conversely, the mirrored on the bisector point is to the previous problem on the circles looking for.

The specified points are denoted by and, the specified straight line with. Furthermore, let the point of intersection of the straight line and with cutting angle. Then, the contact points of the two circles sought after Sekantentangentensatz have the distance of. The centers can be calculated as to the intersections of the bisectors of the perpendicular at the contact points.

Three points

For three points there is a solution. If two points are superimposed, there are infinitely many solutions.

The three points form a triangle with sides a, b, c. The desired circle is the area of this triangle:

For the determination of the area of ​​A is the set of the Heron can be reused.

The problem is solvable in all cases with the classical methods ( compass and straightedge ). If the two groups have in common at least one point M, one can simplify the problem by tracing it back through a reflection in a circle with center M to the case that two of the circles degenerate into straight lines.