Let $c=\mathrm{O}\left(r\right)$ be a fixed circle, and let P be any ordinary point other than O. Let P' be the inverse of P in circle c. Then the line p through P' and perpendicular to OPP' is called the polar of P for the circle c. Since the package does not work with the extended plane, the routine does not find the polar of center O or of the ideal point.

If the point lies on the conic then the polar of P is the tangent line of the conic at that point.

For a detailed description of the polar of P, use the routine detail

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

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

$\mathrm{circle}\left(c\,x^2+y^2=1\,\left[x\,y\right]\right),\mathrm{ellipse}\left(e\,\frac{x^2}{4}+y^2=1\,\left[x\,y\right]\right)\:$

$\mathrm{point}\left(A\,3\,0\right)\:$

$\mathrm{Polar}\left(\mathrm{l1}\,A\,c\right)$

$\mathrm{l1}$

$\mathrm{Equation}\left(\mathrm{l1}\right)$

$-1+3x=0$

$\mathrm{Polar}\left(\mathrm{l2}\,A\,e\right)$

$\mathrm{l2}$

$\mathrm{detail}\left(\mathrm{l2}\right)$

$\begin{array}{ll}\text{name of the object}& \mathrm{l2}\\ \text{form of the object}& \mathrm{line2d}\\ \text{equation of the line}& -1+\frac{3x}{4}=0\end{array}$