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 $(x,y,z)\in P$
On
Without loss of generality, I'll use the plane $x + y + z = 0$ instead of $x + y + z = 1$.
Suppose the circle on the plane $x + y + z = 0$ is a unit circle centered at $(0, 0, 0)$. It intersects the $xy$-plane at $\frac{1}{\sqrt 2}(-1, 1, 0)$ and $\frac{1}{\sqrt 2}(1, -1, 0)$. It intersects the plane $x = y$ at $\frac 1{\sqrt 6}(1, 1, -2)$ and $\frac 1{\sqrt 6}(-1, -1, 2)$. These four points, projected onto the $xy$-plane, define the major and minor axes of the ellipse on the $xy$-plane. (In particular, the major and minor axes have length $2$ and $\frac 2{\sqrt{3}}$, and they make $\pi/4$ angle with the standard $xy$-axis.)
On
Let $C$ be the unit circle on the plane $P$ centered at $R=(0,0,1)\in P$ and $(x_0,y_0)$ be projection of any point of $C$ to $x,y$ plane.
Then $T=(x_0,y_0,1-x_0-y_0)\in C$ and by constraction $d(R,T)=1.$ So, $$|(x_0,y_0,-x_0-y_0)|=1$$ $$x_0^2+y_0^2+x_0^2+2x_0y_0+y_0^2=1$$ $$2x_0^2+2y_0^2+2x_0y_0=1$$ Hence, The equation of ellipse is $2x^2+2y^2+2xy=1$
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 $$ 2 x^2 + 2 xy + 2 y^2 = 1. $$