<s>[INST] Can you find a ellipse so that its image is a circle? This is a "fun" question and I have already a solution. I asked this question so that I may see a different approach or an elegant solution.
Let $P$ be a plane with equation $x+y+z=1$. Find an ellipse on $x,y$ plane so that its "shadow" on $P$ is a circle.
Edit: By "shadow", I mean $(x,y)\to(x,y,z)$ where $(x,y,z)\in P$
 [/INST] The circle with radius 1 and center $(0,0,1)$ in the plane $x+y+z=1$ can be parametrized by
\begin{align}
\vec{p}(t) &= \langle 0,0,1 \rangle + \cos(t)\langle 1,-1,0 \rangle/\sqrt{2} + \sin(t) \langle 1,1,-2 \rangle/\sqrt{6} \\
 &= \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos
   (t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos
   (t)}{\sqrt{2}},1-\sqrt{\frac{2}{3}} \sin (t) \right\rangle.
 \end{align}
It's shadow can be parametrized by just dropping the $z$ coordinate.
$$\vec{s}(t) = \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos
   (t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos
   (t)}{\sqrt{2}} \right\rangle.$$
This is enough to visualize.

Once we have the parametrization, we can generate points on the ellipse by simply plugging in $t$ values.  Here are five points on the ellipse in the plane, for example.
$$
\begin{array}{l|l}
 t & (x,y) \\
\hline
 0 & \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) \\
\hline
 \frac{\pi }{3} & \left(\frac{1}{\sqrt{2}},0\right) \\
\hline
 \frac{\pi }{2} & \left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right) \
\\
\hline
 \pi  & \left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \\
\hline
 \frac{3 \pi }{2} & \left(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)
\end{array}
$$
Since five points determine an ellipse, we can find the Cartesian formula easily enough.  My favorite technique is to set the following determinant equal to zero.
$$
\left|
\begin{array}{cccccc}
 x^2 & x y & y^2 & x & y & 1 \\
 \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} &
   \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \\
 \frac{1}{2} & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 1 \\
 \frac{1}{6} & \frac{1}{6} & \frac{1}{6} &
   \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & 1 \\
 \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} &
   -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \\
 \frac{1}{6} & \frac{1}{6} & \frac{1}{6} &
   -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & 1 \\
\end{array}
\right|
$$
After some simplification, this yields 
$$
2 x^2 + 2 xy + 2 y^2 = 1.
$$
</s>