### Chords in Cardioid

We start with two circles with centers at O (stationary) and P (moving.) Note three points A, M, and T. Initially they coincide. A remains on the stationary circle. M is the point of the moving circle that traces a cardioid. T is the point of contact of the two circles. The fact that the motion is *without slipping* is more accurately expressed as the congruence of the corresponding arcs TA and TM of the two circles.

If at all times OQ is equal and parallel to PM, then Q circles around O at the angular speed which is double that of P (click!). This is then yet another method to draw a cardioid. Build a hinged parallelogram OPMQ with OP twice as long as OQ. Make OQ rotate twice as fast as OP. The fourth vertex M of the parallelogram will trace a cardioid.

A small part of the cardioid can be obtained in an even simpler manner. Consider chords QA. Extend each beyond A to the total length of twice the radius of the circle. As Q moves around the circle, the other end M traces a cardioid. To obtain the whole cardioid, double the length of MQ by extending it beyond Q. The second end, say N, will trace the same cardioid. (For there were two ways to construct the parallelogram to start with.)

Thus we are reminded of an interesting property of the curve: all chord MN through the point A (the *cusp*) of the cardioid have the same length which is 4 times the radius of the generating circles. The midpoint Q of such chords traces a circle which, of course, is not the best way to draw it.

(Click repeatedly in the applet area.)

What if applet does not run? |

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

Copyright © 1996-2018 Alexander Bogomolny

71228903