Answer:
No, but the reason depends. Pick the one that applies to the question.
Are we talking function in two variables, ie [tex]z=f(x,y) = x^2+y^2[/tex] : in this case no, the surface you generate is a paraboloid - think the satellite dishes you use for telecomunications. not linear anyways.
Are we talking about [tex]x^2+y^2=0[/tex]? This has only one solution, which is the origin (0;0). in fact if you solve for either of the two you will get [tex]x^2=-y^2[/tex]. Now LHS is greater or equal than zero, RHS is lesser or equal than zero - the whole thing is true only if they are both 0. But still, not linear.
Are we talking about [tex]x^2+y^2 = r[/tex] with [tex]r>0[/tex] (since if it was we would be again at an absurd, something clearly non-negative is equal to a negative number)? Again, non linear, they are circles of radius [tex]\sqrt r[/tex] centered on the origin.
