# Straight Edge Only Construction of Polar

Given point P and a circle with radius R centered at point O. One needs only a straight edge to construct the polar of P.

### 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. 1. Draw two secants through P. Assume that one intersects the circle at points A and C, the other at points B and D.

2. Let M be the intersection of AD and BC. Let N be the intersection of AB and CD.

3. NM is the desired polar of P.

To justify the construction, we proceed in two steps. First observe that the four lines CDN, ABN, BC, and AD form a complete quadrilateral, for which P serves as the intersection of a pair of diagonals, AC and BD. The latter are cut by the third diagonal MN at points conjugate to P with respect to A, C and B, D, respectfully. Therefore, MN is the polar of P with respect to the circle. Let ROQP be the line through P and the center O; and assume MN meets OP at S. Then S is conjugate of P with respect to R, Q. We have only to show that S is the inverse of P in the given circle. But this is a straightforward property of the inversion.

Since A and B might have been chosen symmetrically with respect to OP, MN is necessarily perpendicular to OP. MN is thus the polar of P with respect to the given circle.

(While toying with the applet, it's instructive to note that moving either A or B does not affect the line MN.)

### References

1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1996
2. D. Wells, The Penguin Dictionary of Curious and Intersting Geometry, Penguin Books, 1991 ## Poles and Polars • 