Straight Edge Only Construction of PolarGiven point P and a circle with radius R centered at point O. One needs only a straight edge to construct the 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
 Poles and Polars
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