Find the orthogonal projection onto xOy of the curve…

Find the orthogonal projection onto $xOy$ of the curve $x^2+y^2-z=0,\ 4x+2y+z+5=0$. What is it?

Explanation of Method

A paraboloid and a plane that meet in a single point: the projection degenerates.

Step 1: Eliminate $z$

\[ x^2+y^2+4x+2y+5=0\iff (x+2)^2+(y+1)^2=0 . \]

\[\text{the single point } (-2,-1,0)\]

Step 2: Meaning

Only $x=-2,\ y=-1$ is possible, and then $z=x^2+y^2=5$. The "curve" is the single point $T(-2,-1,5)$. The plane is tangent to the paraboloid there. The tangent plane of $z=x^2+y^2$ at $T$ is

\[ z=5-4(x+2)-2(y+1)\iff 4x+2y+z+5=0, \] which is exactly the given plane. $\checkmark$

$T(-2,-1,5)$ $(-2,-1)$ $O$

The plane $4x+2y+z+5=0$ (red) touches the paraboloid $z=x^2+y^2$ only at $T(-2,-1,5)$, so the “curve” is a single point and its projection is $(-2,-1)$.

Result

\[\boxed{\,\text{the single point } (-2,-1,0)\,}\]