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Re: Polar Plot


Dear Miguel,
You will probably get some answers about Solve and which 
root to plot, but this may be an instructive example for the 
positive root:

ContourPlot[(r^2 - 3 + 2*r*Cos[theta]) /. {r -> Sqrt[x^2 + y^2],
    theta -> ArcTan[y/x]}, {x, -2, 2}, {y, -2, 2}, Contours -> {0}]


Or better yet:
simpler = Simplify[(r^2 - 3 + 2*r*Cos[theta]) /. {r -> 
Sqrt[x^2 + y^2],
    theta -> ArcTan[y/x]}, {x \[Element] Reals && y \[Element] Reals}]

ContourPlot[simpler, {x, -2, 2}, {y, -2, 2}, Contours -> {0}]


W. Craig Carter

On Mon, 29 Oct 2007, Miguel wrote:

> Date: Mon, 29 Oct 2007 05:36:28 -0500 (EST)
> From: Miguel <misvrne at gmail.com>
> To: mathgroup at smc.vnet.net
> Subject: [mg82709] Polar Plot
> 
> How can I to plot a polar expresion in implicit form. For example,
> r^2==3+2*r*Cos[teta]
>
>
>


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