# Circles with Equal Collinear Chords

Two circles C(O) and C(O'), with centers O and O', and a line L are given. Locate a line parallel to L such that C(O) and C(O') cut off equal chords on L.

In the applet below, the line L is shown at the top of the applet. That line can be dragged as a whole parallel to the previous position or rotated if dragged near the edges of the applet. The two circles can be dragged as a whole, or have their radii changed by dragging their centers.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Two circles C(O) and C(O'), with centers O and O', and a line L are given. Locate a line parallel to L such that C(O) and C(O') cut off equal chords on L.

The problem illustrates an application of the Translation Transform.

It may have 1 or no solutions and, in case the two circles are equal and the line L is parallel to the line of the centers, infinitely many solutions.

Assume the problem solved and let AB and CD be equal chords in C(O) and C(O') on a line parallel to L.

Translate C(O) in the direction of L so as to make AB and CD overlap. The line through O' and the midpoint of segment CD is perpendicular to L and lies on the line parallel to L that passes through O. This leads to the following construction.

Draw a line through O parallel to L and through O' perpendicular,equal,perpendicular,parallel,adjacent to L. Let T be the point of intersection of the two lines. Draw a circle C(T) with center T and the radius equal to that of C(O). If C(T) coincides with C(O') the number of solutions is infinite. If C(T) does not cross C(O'), there are no solutions. Otherwise, let the common chord of C(T) and C(O') be CD and the chord on C(O) obtained by the inverse translation from CD, AB. AB and CD are collinear on a line parallel,equal,perpendicular,parallel,adjacent to L and are equal.

### References

- I. M. Yaglom,
*Geometric Transformations I*, MAA, 1962, Problem 8(a)

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

63602202 |