Revisit "How can I visualize the nuclear norm ball" Revisiting How can I visualize the nuclear norm ball?

Two eigenvalues are reproduced as following:
$$ s_{1,2}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2\pm|x+z|\sqrt{(x-z)^2+4y^2}}. $$
According to the following (from a paper)

If a symmetric matrix: $$ A=\left( \begin{array}{cc} x & y\\ y & z\end{array} \right)$$
is rank $1$, then $y=\sqrt{xz}$, which comes from the fact that $vv^T$ is rank $1$ and any rank $1$ matrix can be represented in this form.
$$\left[\begin{array}{cc} v_1\\ v_2\end{array}\right]\left[\begin{array}{cc} v_1 & v_2\end{array}\right]=\left( \begin{array}{cc} v_1^2 & v_1v_2\\ v_1v_2 & v_2^2\end{array} \right)$$

My question: how to explain the red circle in figure (b) is the $2\times 2$ symmetric unit-Euclidean-norm rank $1$ matrix?
This is a circle in $3$-D, how to get the equation of this circle through the rank $1$ matrix provided above?

I believe just replace $y=\pm\sqrt{xz}$ in $s_{1,2}$ and can get the answer.
So I choose the larger one of $s_{1,2}$:
$$ s_{\max}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2 + |x+z|\sqrt{(x-z)^2+4y^2}}= \sqrt{2(x+z)^2}=1 (\text{unit norm})$$

$A$ must be rank-$1$ and symmetric. Thus
$$A=vv^T$$
This implies that
$$y^2=xz$$
In this problem, we also have $\|v\|=1$, so
$$x+z=1$$
All of these result in
$$y=\pm\sqrt{x(1-x)},\quad z=1-x;$$
