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This question already has an answer here:

Consider:

f = Sqrt[9 - x^2 - y^2];
ContourPlot[f, {x, -4, 4}, {y, -4, 4},
 Contours -> {0, 1, 2, 3},
 ContourLabels -> All,
 ContourShading -> None]

Which produces this image:

enter image description here

Now, I understand that the contour f=3 is just the point (0,0) and maybe that's why it's not drawn. But I don't understand why the contour f=0, which should be a circle of radius 3, is missing?

Any thoughts?

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marked as duplicate by Michael E2 plotting Jul 9 '15 at 5:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ For x=y=3, f is complex $\endgroup$ – Enrique Pérez Herrero Jul 8 '15 at 19:20
  • $\begingroup$ @EnriquePérezHerrero 9 - x^2 - y^2 /. {x -> 3/Sqrt@2, y -> 3/Sqrt@2} $\endgroup$ – Dr. belisarius Jul 8 '15 at 19:23
  • $\begingroup$ Sqrt[9 - 3^2 - 3^2] = 3i $\endgroup$ – Enrique Pérez Herrero Jul 8 '15 at 19:25
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    $\begingroup$ @EnriquePérezHerrero: When finding the contour $f=0$, we're working with $\sqrt{9-x^2-y^2}=0$, which is equivalent to $x^2+y^2=9$, which is a circle of radius 3. We are not substituting 3 for $x$ and $y$. But see Michael Seifert's answer. But thanks for the reply. $\endgroup$ – David Jul 8 '15 at 20:05
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    $\begingroup$ Possible duplicates: (23363), (32734) $\endgroup$ – Michael E2 Jul 8 '15 at 22:59
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It appears to be a problem with the square root:

f = 9 - x^2 - y^2;
ContourPlot[f, {x, -4, 4}, {y, -4, 4}, Contours -> {0, 1, 4, 9}, 
 ContourLabels -> (Text[Sqrt[#3], {#1, #2}] &), ContourShading -> None]

enter image description here

My guess is that the algorithm used by ContourPlot implicitly relies on the function having a smooth gradient in some neighborhood of each contour, which the function $f = \sqrt{9 - x^2 - y^2}$ does not have in the neighborhood of the level set $f = 0$.

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  • $\begingroup$ Probably a good point. Thanks for the help. $\endgroup$ – David Jul 8 '15 at 19:53
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f = Sqrt[9 - x^2 - y^2];

Reduce[f == 0, {x, y}, Reals]

-3 <= x <= 3 && (y == -Sqrt[9 - x^2] || y == Sqrt[9 - x^2])

Mathematica has a hard time finding a real solution for f == 0. Using a large number of PlotPoints and forcing the result to be real using Re produces a highly segmented plot (multiple labels for 0)

ContourPlot[Re[f], {x, -4, 4}, {y, -4, 4}, Contours -> {0, 1, 2, 3},
 ContourLabels -> All,
 ContourShading -> None,
 PlotPoints -> 201]

enter image description here

For a workaround use RegionPlot:

Row[{
  RegionPlot[
   ImplicitRegion[
      Reduce[
       Sqrt[9 - x^2 - y^2] == #, {x, y}, Reals],
      {x, y}] & /@
    Range[0, 3],
   PlotRange -> {{-4, 4}, {-4, 4}},
   ImageSize -> 360],
  LineLegend[
   {Red, Darker[Green], Orange, Blue},
   Range[3, 0, -1]]}]

enter image description here

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