# Find a Common Chord of Given Length

Given two circles C(O_{1}) and C(O_{2}), with centers O_{1} and O_{2}, respectively, intersecting at points P and Q. Construct a line through P, such that it intersects C(O_{1}) and C(O_{2}) in two other (than P) points M_{1} and M_{2} so that the segment M_{1}M_{2} 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.

What if applet does not run? |

### References

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

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Copyright © 1996-2018 Alexander Bogomolny

Given two circles C(O_{1}) and C(O_{2}), with centers O_{1} and O_{2}, respectively, intersecting at points P and Q. Construct a line through P, such that it intersects C(O_{1}) and C(O_{2}) in two other (than P) points M_{1} and M_{2} so that the segment M_{1}M_{2} has a given length a.

Assume that the problem is solved and points M_{1} on C(O_{1})) and M_{2} on C(O_{2}) are such that M_{1}M_{2} = a. Let C_{1} and C_{2} be the feet of the perpendiculars from O_{1} and O_{2} on the line M_{1}M_{2}. Since C_{1} is the midpoint of M_{1}P and C_{2} the midpoint of M_{2}P, _{1}C_{2} = a/2._{1}C_{2} so as to make C_{1} coincide with O_{1}. Let the new position of C_{2} be denoted T.

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Then ΔO_{1}O_{2}T is a right triangle with hypotenuse O_{1}O_{2} and _{1}T = a/2.

Form a circle C(O) with diameter O_{1}O_{2} and another _{1}, a/2),_{1} 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(O_{1}, a/2). Then _{1}T = a/2,_{1}T cuts off circles C(O_{1}) and C(O_{2}) a common chord M_{1}M_{2} of the given length a.

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Copyright © 1996-2018 Alexander Bogomolny

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