<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] There is an error in your algebra
$$\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac32$$ 
so parametrically continue from here.
$$ x = \sqrt{\frac32} \cos t -\frac12\,,y =\sqrt{\frac32}  \sin t -\frac12 $$
Plug these into the second equation and find $z$ in terms of $t$.
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