Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$ 
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$.

So if we substitute $z=1-x-y$ into the equation $x^2+y^2=z$ we get $x^2+y^2=1-x-y$ which can be written in the form $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=1$. 
Is this the curve of intersection? Because it seems to me (by geometrical intuition) that the curve should be an ellipse. What is the logical flaw here. Help please!

After the coments, a possible parametrization is
$$r(t)=\left(\,-\frac12+\sqrt\frac32\,\cos t\,,\,\,-\frac12+\sqrt\frac32\,\sin t\,,\,\,2-\sqrt\frac32(\cos t+\sin t)\,\right)\;,\;\;0\le t\le 2\pi$$
Observe that we have above a general vector of the form $\;(x,y,z)\;$ , where for the third coordinate we have either $\;z=x^2+y^2\;$ or $\;z=1-x-y\;$ .
