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I discovered that Mac's Grapher has a parametric polar mode, i.e. where $r$ and $\theta$ can be specified in terms of a parameter, usually $t$. I am attempting to convert the generic equation for a circle into this format, i.e. from $(x-h)^2+(y-k)^2=g^2$. (I randomly chose $g$ as the radius to avoid $r$ collisions.) It is not going well. I keep making cardioids or weirder shapes. How can you specify equations for $r$ and $\theta$ in terms of $t$ for a circle of radius $g$ and Cartesian center of $(h,k)$?

attempt so far [http://imgur.com/0iNOpqR ]

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  • $\begingroup$ en.wikipedia.org/wiki/Polar_coordinate_system#Circle gives the general equation of a circle in polar coordinates. Parametrize away! If you want to work it out for yourself, I suggest starting with the cartesian equation and converting it to polar. $\endgroup$ – amd May 30 '16 at 17:03
  • $\begingroup$ Isn't that just $x=h+rcos(t)$ and $y=k+rsin(t)$? $\endgroup$ – Robert Marshall Murphy May 30 '16 at 17:16
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One possibility (not very pretty) is to take a parametric form of the cartesian equation and convert that to polar coordinates. Starting from your suggested $x=h+g\cos t$, $y=k+g\sin t$ you can get $$\begin{align} r&=\sqrt{(h+g\cos t)^2+(k+g\sin t)^2} \\ \theta &= \arctan{k+g\sin t\over h+g\cos t}.\end{align}$$ You’ll need to use the atan2 function in Grapher so that $\theta$ ends up in the correct quadrant.

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  • $\begingroup$ I fear that is as good as it gets. The polar form ($r=...$) of a circle $g$ about $(h,k)$ is pretty nasty, so that makes sense. You have my vote! $\endgroup$ – Robert Marshall Murphy May 30 '16 at 18:33
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A cross between polar and cartesian will lead you nowhere.

A two parameter annular ring of eccentric circles of radius $R$ is:

$$ x = \sqrt { e^2 + R^2 + 2 e R \cos \phi} \, \cos \theta ; \, y= \sqrt { e^2 + R^2 + 2 e R \cos\phi} \, \sin \theta , $$

out of which choose a single value of desired polar angle $\theta$ for radius vector of circle centre.

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