Find a Common Chord of Given Length

Given two circles C(O₁) and C(O₂), with centers O₁ and O₂, respectively, intersecting at points P and Q. Construct a line through P, such that it intersects C(O₁) and C(O₂) in two other (than P) points M₁ and M₂ so that the segment M₁M₂ has a given length a.

In the applet below, at the top of the applet, there is a line segment with changeable end points that signifies the given length a. The two circles can be dragged as a whole, or have their radii changed by dragging their centers.

Solution

References

1. I. M. Yaglom, Geometric Transformations I, MAA, 1962, Problem 7(a)

Copyright © 1996-2010 Alexander Bogomolny

Assume that the problem is solved and points M₁ on C(O₁)) and M₂ on C(O₂) are such that M₁M₂ = a. Let C₁ and C₂ be the feet of the perpendiculars from O₁ and O₂ on the line M₁M₂. Since C₁ is the midpoint of M₁P and C₂ the midpoint of M₂P, C₁C₂ = a/2. Translate C₁C₂ so as to make C₁ coincide with O₁. Let the new position of C₂ be denoted T.

Then ΔO₁O₂T is a right triangle with hypotenuse O₁O₂ and O₁T = a/2. This suggests the following construction:

Form a circle C(O) with diameter O₁O₂ and another C(O₁, a/2), with center O₁ and radius a/2. If the two intersect or are tangent to each other, the problem is solvable. It has 2 solutions in the former case and one solution,one solution,two solutions,three solutions in the latter case. Otherwise, it has no solutions.

Let T be the common point of C(O) and C(O₁, a/2). Then O₁T = a/2, and the line through P parallel,equal,perpendicular,parallel,adjacent to O₁T cuts off circles C(O₁) and C(O₂) a common chord M₁M₂ of the given length a.

Copyright © 1996-2010 Alexander Bogomolny