The curve c is the intersection of the sphere…
The curve $c$ is the intersection of the sphere $x^2+y^2+z^2=4(x-2y-2z)$ with the plane through its center normal to the line $x=0,\ y=-z$. Find the projection of $c$ onto $z=0$.
Explanation of Method
A great circle of a sphere and its projection onto xOy, an ellipse.
Step 1: Sphere and plane
\[ (x-2)^2+(y+4)^2+(z+4)^2=36,\qquad C(2,-4,-4),\ R=6 . \]
The line has direction $\{0,1,-1\}$. The plane through $C$ with this normal is $(y+4)-(z+4)=0$, that is $y=z$. It passes through $C$, so $c$ is a great circle of radius $6$.
Step 2: Projection
Eliminate $z=y$: $(x-2)^2+2(y+4)^2=36,\quad z=0$.
\[ \frac{(x-2)^2}{36}+\frac{(y+4)^2}{18}=1 . \]
This is an ellipse with center $(2,-4,0)$ and semi-axes $6$ and $3\sqrt2$. The plane $y=z$ makes $45^\circ$ with $xOy$, so the $y$-direction shrinks by $\cos45^\circ$: $6\cdot\tfrac{\sqrt2}2=3\sqrt2$.
Result
\[\boxed{\,(x-2)^2+2(y+4)^2=36,\quad z=0\,}\]