= Solution
The stationary equations are
$$
2x(y-a)=0,
\qquad x^2+y^2=1.
$$
Thus
$$
P_+=(0,1),
\qquad P_-=(0,-1),
$$
always exist, and for $|a|\leq1$ there are also
$$
Q_\pm=(\pm\sqrt{1-a^2},a).
$$
The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}2(y-a)&2x\\-2x&-2y\end{pmatrix}.
$$
At $P_+$ its eigenvalues are $2(1-a),-2$, so it is a saddle for $a<1$, a stable node for $a>1$, and nonhyperbolic at $a=1$. At $P_-$ they are $-2(1+a),2$, so it is a saddle for $a>-1$, an unstable node for $a<-1$, and nonhyperbolic at $a=-1$.
At $Q_\pm$, for $|a|<1$, the characteristic polynomial is
$$
\lambda^2+2a\lambda+4(1-a^2).
$$
The equilibria are stable for $a>0$ and unstable for $a<0$; they are foci when $|a|<2/\sqrt5$ and nodes when $2/\sqrt5<|a|<1$, with a repeated-eigenvalue transition at equality. For $a=0$ the eigenvalues are purely imaginary. In that case the system is Hamiltonian with <first integral>
$$
H(x,y)=xy^2+\frac{x^3}{3}-x,
$$
because $\dot x=H_y$ and $\dot y=-H_x$. The Hessian of $H$ is positive definite at $(1,0)$ and negative definite at $(-1,0)$, so nearby regular level sets are closed curves. Hence both nonhyperbolic equilibria are genuine nonlinear centers.
Finally,
$$
\nabla\mathbin\cdot(\dot x,\dot y)
=2(y-a)-2y=-2a.
$$
For $a\ne0$ this has a strict constant sign throughout the simply connected plane, so the <Bendixson-Dulac criterion> proves that \b[there are no periodic orbits]. These results comprise the <phase portrait of x dot equals two x times y minus a>.
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