Let I and E be the internal and external centers of similitude of two given nonconcentric circles c1 and c2 having unequal radii. Then the circle on IE as diameter is called the circle of similitude of the two given circles.

If the third optional is given and is of the form 'centername' = cn where cn is name, cn will be the name of the center of c.

For a detailed description of the center of similitude c, use the routine detail (i.e., detail(c))

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

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

$\mathrm{\_EnvHorizontalName}≔'x'\:$$\mathrm{\_EnvVerticalName}≔'y'\:$

$\mathrm{circle}\left(\mathrm{c1}\,x^2+y^2=1\,'\mathrm{centername}'=\mathrm{o1}\right)\:$

$\mathrm{circle}\left(\mathrm{c2}\,\left[\mathrm{point}\left(A\,3\,3\right)\,4\right]\,'\mathrm{centername}'=\mathrm{o2}\right)\:$

$\mathrm{CircleOfSimilitude}\left(c\,\mathrm{c1}\,\mathrm{c2}\,'\mathrm{centername}'=o\right)$

$c$

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

$\begin{array}{ll}\text{name of the object}& c\\ \text{form of the object}& \mathrm{circle2d}\\ \text{name of the center}& o\\ \text{coordinates of the center}& \left[-\frac{1}{5}\,-\frac{1}{5}\right]\\ \text{radius of the circle}& \frac{\sqrt{128}\sqrt{25}}{50}\\ \text{equation of the circle}& -\frac{6}{5}+x^2+y^2+\frac{2}{5}x+\frac{2}{5}y=0\end{array}$

$\mathrm{draw}\left(\left\{c\,\mathrm{c1}\,\mathrm{c2}\right\}\,\mathrm{printtext}=\mathrm{true}\,\mathrm{view}=\left[-2..4\,-2..4\right]\right)$