# 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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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Solution

### References

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

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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

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. 