Let $c=\mathrm{O}\left(r\right)$ be a fixed circle and let P be any ordinary point other than the center O. Let P' be the inverse of P in circle $\mathrm{O}\left(r\right)$. Then the line Q through P' and perpendicular to OPP' is called the polar of P for the circle c.  Note that when P is a line, then Q will be a point.

Note that this routine in particular, and the geometry package in general, does not encompass the extended plane, i.e., the polar of center O does not exist (though in the extended plane, it is the line at infinity) and the polar of an ideal point P does not exist either (it is the line through the center O perpendicular to the direction OP in the extended plane).

If line Q is the polar point P, then point P is called the pole of line Q.

The pole-polar transformation set up by circle $c=\mathrm{O}\left(r\right)$ is called reciprocation in circle c

For a detailed description of Q (the object created), use the routine detail (i.e., detail(Q))

The command with(geometry,reciprocation) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{geometry}\right)\:$

$\mathrm{circle}\left(c\,\left[\mathrm{point}\left(\mathrm{OO}\,0\,0\right)\,2\right]\,\left[x\,y\right]\right)\:$

$\mathrm{point}\left(P\,1\,0\right)\:$

$\mathrm{inversion}\left(\mathrm{PP}\,P\,c\right)\:$

$\mathrm{reciprocation}\left(l\,P\,c\right)$

$l$

$\mathrm{detail}\left(l\right)$

$\begin{array}{ll}\text{name of the object}& l\\ \text{form of the object}& \mathrm{line2d}\\ \text{equation of the line}& -4+x=0\end{array}$

$\mathrm{draw}\left(\left\{c\,l\,\mathrm{dsegment}\left(\mathrm{dsg}\,P\,\mathrm{PP}\right)\right\}\,\mathrm{printtext}=\mathrm{true}\,\mathrm{view}=\left[-3..5\,-3..3\right]\,\mathrm{axes}=\mathrm{NONE}\right)$

$\mathrm{reciprocation}\left(A\,l\,c\right)\:$

$\mathrm{coordinates}\left(A\right)=\mathrm{coordinates}\left(P\right)$

$\left[1\,0\right]=\left[1\,0\right]$