Imaginary variables in ImplicitRegion

I am using the function ImplicitRegion[].

RegionPlot[ImplicitRegion[x^2 + y^2 == -1, {x, y}]] which gives me a circle, though in the documentation variables are said to be real.

RegionPlot[ Assuming[{x, y} \[Element] Reals,
ImplicitRegion[   x^2 + y^2 == -1, {{x, -\[Pi], \[Pi]}, {y, -\[Pi], \[Pi]}}]]]

How to fix this issue?

If ImplicitRegion was working properly you would not get a plot. If you are trying to plot the equation:

funcs = y /. Solve[x^2 + y^2 == -1, y];

ReImPlot[funcs, {x, -Pi, Pi},
PlotStyle -> {AbsoluteThickness, Automatic},
PlotLegends -> Automatic] • But is there a way to tell ImplicitRegion that (x,y) are real and thus there are no solutions? I am using this function in NIntegrate so it's important not to get fake contours. Apr 8 '21 at 6:07
• ImplicitRegion does not take an Assumptions option so you have to explicitly include the domain constraint. RegionPlot[ImplicitRegion[x^2 + y^2 == -1 && Element[{x, y}, Reals], {x, y}], {x, -Pi, Pi}, {y, -Pi, Pi}] will return an empty plot. Apr 8 '21 at 6:15
• I tried changing -1 to 1 in your example and it gave an empty plot again. Apr 8 '21 at 7:57
• Yes; however, you indicated that your actual problem is integration. Both Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == 1, {x, y}]] and Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == -1, {x, y}]] return the expected results. As do Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == 1 && Element[{x, y}, Reals], {x, y}]] and Integrate[1, {x, y} \[Element] ImplicitRegion[x^2 + y^2 == -1 && Element[{x, y}, Reals], {x, y}]] Apr 8 '21 at 13:00
• oh, I see, thanks! Perhaps the problem is related to RegionPlot. Apr 8 '21 at 15:43