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I have a function of the type: 

F[{x_,y_}] := x^2 + y^2

One can plot this function using Plot3D:

Plot3D[F[{x,y}], {x, -2, 2}, {y, -2, 2}]

If one plots the function, it is clear that F[{x,y}] = 6 at more than one set of coordinates.

How can I obtain all sets of coordinates for which F[{x,y}] = 6?

Thanks, Stuart.

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2 Answers 2

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Well there is a nice way to extract points:

mYcp = ContourPlot[x^2 + y^2 == 6, {x, -3, 3}, {y, -3, 3}]

enter image description here

Short[Cases[Normal@mYcp, Line[x_] :> x, Infinity], 20]

{{{-1.07143,-2.20209},{-1.1106,-2.18203},{-1.17857,-2.14721},{-1.18634,-2.14286},{-1.24885,-2.10599},{-1.28571,-2.08271},{-1.38018,-2.02304},{-1.47143,-1.95714},{-1.5,-1.93609},{-1.50433,-1.9329},{-1.50952,-1.92857},{-1.62121,-1.8355},{-1.71429,-1.7479},{-1.7316,-1.7316},{-1.7479,-1.71429},<<246>>,{-0.289216,-2.43207},{-0.321429,-2.42778},{-0.363553,-2.42216},{-0.383423,-2.41914},{-0.428571,-2.41111},{-0.473545,-2.40212},{-0.634454,-2.36555},{-0.642857,-2.36335},{-0.647783,-2.36207},{-0.663265,-2.35714},{-0.857143,-2.29252},{-0.961538,-2.25275},{-0.965438,-2.25115},{-0.968045,-2.25},{-1.07143,-2.20209}}}

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  • $\begingroup$ wow great ! thanks for sharing it, +1 $\endgroup$
    – Ali Hashmi
    Mar 25, 2016 at 10:52
  • $\begingroup$ Nice, thanks for the solution. $\endgroup$
    – user14424
    Mar 25, 2016 at 11:01
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Perhaps not a good job but the closest i can get to:

plot = Reap[ContourPlot[F[{x, y}], {x, -3, 3}, {y, -3, 3},
EvaluationMonitor :> If[5.95 < F[{x, y}] < 6, Sow[{x, y, F[{x, y}]}],]]][[2,1]] // 
#[[All, {1, 2}]] & //ListPlot[#, PlotStyle -> Red] &

enter image description here

Show[ContourPlot[F[{x, y}] == 6, {x, -3, 3}, {y, -3, 3}], plot]

enter image description here

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  • $\begingroup$ There are certainly numerical errors associated with the way i extracted points from the plot. however, if accuracy is not a big issue, the answers will be quite reasonable $\endgroup$
    – Ali Hashmi
    Mar 25, 2016 at 8:55
  • $\begingroup$ Hi Ali, thanks for your answer. The example I posted was a much simpler function that the one I am actually dealing with. I have now tried your approach on my function and the results don't look too bad. Do you have any more ideas on how to improve the accuracy? Thanks, Stuart. $\endgroup$
    – user14424
    Mar 25, 2016 at 9:26
  • $\begingroup$ Or alternatively, incorporate more points? $\endgroup$
    – user14424
    Mar 25, 2016 at 9:27
  • $\begingroup$ search for the question titled, "getting a list of accurate coordinates from a plot". I think someone used a mouse event to get all the coordinates by hovering the mouse over the plot. It may be more reasonable $\endgroup$
    – Ali Hashmi
    Mar 25, 2016 at 9:37
  • $\begingroup$ Hi Ali, thanks for the help again. Looking back at your previous answer the contour plot is completely sufficient. Thanks for your help. Cheers, Stuart. $\endgroup$
    – user14424
    Mar 25, 2016 at 10:31

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