reciprocation - Maple Help

geometry

 reciprocation
 find the reciprocation of a point or a line with respect to a circle

 Calling Sequence reciprocation(Q, P, c)

Parameters

 Q - the name of the object to be created P - point or line c - circle

Description

 • 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.

Examples

 > $\mathrm{with}\left(\mathrm{geometry}\right):$
 > $\mathrm{circle}\left(c,\left[\mathrm{point}\left(\mathrm{OO},0,0\right),2\right],\left[x,y\right]\right):$
 > $\mathrm{point}\left(P,1,0\right):$
 > $\mathrm{inversion}\left(\mathrm{PP},P,c\right):$
 > $\mathrm{reciprocation}\left(l,P,c\right)$
 ${l}$ (1)
 > $\mathrm{detail}\left(l\right)$
 $\begin{array}{ll}{\text{name of the object}}& {l}\\ {\text{form of the object}}& {\mathrm{line2d}}\\ {\text{equation of the line}}& {-}{4}{+}{x}{=}{0}\end{array}$ (2)
 > $\mathrm{draw}\left(\left\{c,l,\mathrm{dsegment}\left(\mathrm{dsg},P,\mathrm{PP}\right)\right\},\mathrm{printtext}=\mathrm{true},\mathrm{view}=\left[-3..5,-3..3\right],\mathrm{axes}=\mathrm{NONE}\right)$
 > $\mathrm{reciprocation}\left(A,l,c\right):$
 > $\mathrm{coordinates}\left(A\right)=\mathrm{coordinates}\left(P\right)$
 $\left[{1}{,}{0}\right]{=}\left[{1}{,}{0}\right]$ (3)