If I have the equation (x-2)^2 + (y-2)^2 == 1 I can use a ContourPlot, however am having trouble changing the axes origin. I have tried doing this:

Plot[y /. Solve[(x-2)^2+(y-2)^2], {x,0,3}, AspectRatio -> Automatic, AxesOrigin -> {0, 0}]

However the top and bottom of the circle are slightly separated: enter image description here

Also how would I use RevolutionPlot3D to plot equation around the Y axis ?

Please let me know if you need more information.


1 Answer 1


For me, Plot[y/.Solve[(x-2)^2+(y-2)^2]==1,{x,0,3}, AspectRatio->Automatic, AxesOrigin->{0,0}] did not have noticeable separation so I cannot help with that. I also successfully used ContourPlot if that works better for you. ContourPlot[(x - 2)^2 + (y - 2)^2, {x, -2, 4}, {y, -2, 4}, Contours -> {1}]

If you just need circles, consider looking into Circle, Sphere, and Ball, which are very simple to use with Graphicsand Graphics3D

If you want a sphere with RevolutionPlot3D, I think you will need to plot it one half at a time

Show[RevolutionPlot3D[Sqrt[1 - (x)^2], {x, 0, 1}, {t, 0, 2 \[Pi]}], 
     RevolutionPlot3D[-Sqrt[1 - (x)^2], {x, 0, 1}, {t, 0, 2 \[Pi]}], 
     PlotRange -> All, AspectRatio -> 1]

circle rotated

I assumed you wanted the non shifted version of the sphere rotated. The shifted version would work the same though:

Show[RevolutionPlot3D[Sqrt[1 - (x - 2)^2] + 2, {x, 1, 3}, {t, 0, 2 \[Pi]}], 
     RevolutionPlot3D[-Sqrt[1 - (x - 2)^2] + 2, {x, 1, 3}, {t, 0, 2 \[Pi]}],   
     PlotRange -> All]

shifted circle rotated

  • $\begingroup$ Thanks the revolution plot works perfectly. I've added a picture of what happens when I do Plot[y/.Solve[(x-2)^2+(y-2)^2]==1,{x,0,3}, AspectRatio->Automatic, AxesOrigin->{0,0}]. Also ContourPlot does not allow me to change the Axes origin like it does with the plot function. $\endgroup$ Commented Jul 29, 2015 at 6:38

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