If P is a point that is not the same as the center O of circle $c\left(r\right)$, the inverse of P in, or with respect to, circle $c\left(r\right)$ is the point Q lying on the line OP such that $\mathrm{SensedMagnitude}\left(\mathrm{OP}\right)\mathrm{SensedMagnitude}\left(\mathrm{OQ}\right)=r^2$.

If P is a line passing through center O of circle $c\left(r\right)$, the inverse of P is P itself. In case P is a line not passing through center O of circle $c\left(r\right)$, the inverse of P is a circle Q passing though O perpendicular to P

If P is a circle passing through the center O of circle $c\left(r\right)$, the inverse of P is a straight line Q not passing through O and perpendicular to the diameter of $c\left(r\right)$ through O. In case P is a line not passing through the center O of circle $c\left(r\right)$, the inverse of P is a circle Q not passing through O and homothetic to circle $c\left(r\right)$ with O as center of homothety.

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

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

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

$\mathrm{point}\left(A\,2\,0\right)\:$$\mathrm{circle}\left(\mathrm{c1}\,x^2+y^2=16\,\left[x\,y\right]\right)\:$

$\mathrm{inversion}\left(B\,A\,\mathrm{c1}\right)\:$$\mathrm{inversion}\left(C\,B\,\mathrm{c1}\right)\:$

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

$\left[2\,0\right]=\left[2\,0\right]$

$\mathrm{line}\left(\mathrm{l1}\,y=x\,\left[x\,y\right]\right)\:$

$\mathrm{IsOnLine}\left(\mathrm{center}\left(\mathrm{c1}\right)\,\mathrm{l1}\right)$

$\mathrm{true}$

$\mathrm{inversion}\left(\mathrm{l2}\,\mathrm{l1}\,\mathrm{c1}\right)\:$

$\mathrm{Equation}\left(\mathrm{l1}\right)=\mathrm{Equation}\left(\mathrm{l2}\right)$

$\left(y-x=0\right)=\left(y-x=0\right)$

$\mathrm{line}\left(k\,x=2\,\left[x\,y\right]\right)\:$

$\mathrm{inversion}\left(\mathrm{k1}\,k\,\mathrm{c1}\right)\:$$\mathrm{inversion}\left(\mathrm{kk1}\,\mathrm{k1}\,\mathrm{c1}\right)\:$

$\mathrm{form}\left(\mathrm{k1}\right)$

$\mathrm{circle2d}$

$\mathrm{Equation}\left(k\right),\mathrm{Equation}\left(\mathrm{kk1}\right)$

$x-2=0,-16+8x=0$

$\mathrm{circle}\left(\mathrm{c2}\,\left[\mathrm{point}\left(A\,4\,0\right)\,1\right]\,\left[x\,y\right]\right)\:$

$\mathrm{IsOnCircle}\left(\mathrm{center}\left(\mathrm{c2}\right)\,\mathrm{c1}\right)$

$\mathrm{true}$

$\mathrm{inversion}\left(\mathrm{c3}\,\mathrm{c2}\,\mathrm{c1}\right)\:$

$\mathrm{form}\left(\mathrm{c3}\right)$

$\mathrm{circle2d}$

$\mathrm{circle}\left(\mathrm{c2}\,{\left(x-3\right)}^2+y^2=36\,\left[x\,y\right]\right)\:$

$\mathrm{inversion}\left(\mathrm{c3}\,\mathrm{c1}\,\mathrm{c2}\right)\:$$\mathrm{inversion}\left(\mathrm{c4}\,\mathrm{c3}\,\mathrm{c2}\right)\:$

$\mathrm{Equation}\left(\mathrm{c1}\right)=\mathrm{Equation}\left(\mathrm{c4}\right)$

$\left(x^2+y^2-16=0\right)=\left(x^2+y^2-16=0\right)$

$\mathrm{detail}\left(\left[\mathrm{c1}\,\mathrm{c2}\,\mathrm{c3}\right]\right)$

$\left[\begin{array}{ll}\text{name of the object}& \mathrm{c1}\\ \text{form of the object}& \mathrm{circle2d}\\ \text{name of the center}& \mathrm{center\_c1}\\ \text{coordinates of the center}& \left[0\,0\right]\\ \text{radius of the circle}& \sqrt{16}\\ \text{equation of the circle}& x^2+y^2-16=0\end{array}\,\begin{array}{ll}\text{name of the object}& \mathrm{c2}\\ \text{form of the object}& \mathrm{circle2d}\\ \text{name of the center}& \mathrm{center\_c2}\\ \text{coordinates of the center}& \left[3\,0\right]\\ \text{radius of the circle}& \sqrt{36}\\ \text{equation of the circle}& x^2+y^2-6x-27=0\end{array}\,\begin{array}{ll}\text{name of the object}& \mathrm{c3}\\ \text{form of the object}& \mathrm{circle2d}\\ \text{name of the center}& \mathrm{center\_c3}\\ \text{coordinates of the center}& \left[\frac{129}{7}\,0\right]\\ \text{radius of the circle}& -\frac{36\sqrt{16}}{7}\\ \text{equation of the circle}& \frac{49}{1296}{x}^2-\frac{301}{216}x-\frac{455}{144}+\frac{49}{1296}{y}^2=0\end{array}\right]$