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!

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 $t$.
