# Plotting 2D function and other 3D functions coincidently

ContourPlot3D[{x^2 + y^2 == z, 2 y == z, x^2 + (y - 1/2)^2 == 1}, {x, -2, 2}, {y, -0.5, 1}, {z, 0, 2}]

The above isn't what I'm seeking. I'd like to plot $x^2 + (y - 1/2)^2 = 1$ in 2D, so as a circle on the xy-plane ($z = 0$), in the same graph as the other two functions that I'd like in 3D. How would I do this?

halirutan's answer nearly resolves this, but his plot still contains $x^2 + (y - 1/2)^2 = 1$ in 3D for all $z$.
I want solely the trace in the xy-plane, for the combined plot of all 3 functions.

• Do you mean you want to get rid of the 3D surface $x^2 + (y - 1/2)^2 = 1$ in halirutan's plot? Just remove it from the ContourPlot3D: i.stack.imgur.com/TKsvo.png (although I prefer i.stack.imgur.com/OVJ3R.png). – user484 Apr 26 '14 at 22:09

You question is not completely clear. What you can do is the following: You make a 2d contour plot of your second equation and transform this into a 3d graphics by appending a 0 (z=0) to all Line points appearing in the 2d graphics and turning the lines into Tube's:

Show[
ContourPlot3D[{x^2 + y^2 == z, 2 y == z,
x^2 + (y \[Minus] 1/2)^2 == 1}, {x, -2, 2}, {y, -0.5, 1}, {z, 0,
2}, ContourStyle -> Opacity[.3]],
Graphics3D[{Red,
Cases[Normal[
ContourPlot[x^2 + (y - 1/2)^2 == 1, {x, -2, 2}, {y, -1/2, 1}]],
Line[pts_] :> Tube[Append[#, 0] & /@ pts], Infinity]}],
PlotRange -> All
] • +1. Thanks. I emended my question. Do you seize it now? – Accounting Apr 4 '14 at 6:18
• @LePressentiment So you don't want a combined plot? Your 2d graphics can be obtained by ContourPlot[x^2 + (y - 1/2)^2 == 1, {x, -2, 2},{y, -1/2, 1}]]. The question remains, what do you want to do with this 2d graphic? – halirutan Apr 4 '14 at 6:57
• Thanks. I do want a combined plot. I've emended my question again. Does this answer your question: "The question remains, what do you want to do with this 2d graphic?" – Accounting Apr 5 '14 at 15:04