<s>[INST] 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!
 [/INST] 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 $\;z=1-x-y\;$ .
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