Now let $k$ be a finite field. If $E$ is an elliptic curve over $k$ then $End(E)\otimes\mathbf{Q}$ is either an imaginary quadratic field or a definite quaternion algebra over $\mathbf{Q}$.

Warning: $E$ may be supersingular even if $End(E)\otimes\mathbf{Q}$ is an imaginary quadratic field; this means that not all endomorphisms of $E$ are defined over $k$.

Again, if $A$ is not $k$-simple then it is isogenous either to a product $E_1\times E_2$ of two mutually non-isogenous elliptic curves $E_1$ and $E_2$ over $k$ (and $\rho(A)=2$) or to a square $E^2$ of an elliptic curve $E$ over $K$ (and $\rho(A)=4$ or $6$). More precisely, $\rho(A)=6$ if and only if $A$ is isogenous to $E^2$ where $E$ is a supersingular elliptic curve, all whose endomorphisms are defined over $k$.
(Indeed, it may happen that both $E_1$ and $E_2$ are supersingular but not isogenous over $k$ and then $\rho(A)=2$)

If $A$ is simple then one of the following conditions holds.

I) $End(A)\otimes \mathbf{Q}$ is a CM-field of degree $4$ and $\rho(A)=2$.

2) $End(A)\otimes \mathbf{Q}$ is a quaternion algebra over an imaginary quadratic field. Then $A$ is supersingular but not all its endomorphisms are defined over $k$ and $\rho(A)=4$.

3) $End(A)\otimes \mathbf{Q}$ is a totally definite quaternion algebra over a real quadratic field. Then $A$ is supersingular but not all its endomorphisms are defined over $k$ and $\rho(A)=2$. (See Tate's Bourbaki talk http://archive.numdam.org/ARCHIVE/SB/SB_1968-1969__11_/SB_1968-1969__11__95_0/SB_1968-1969__11__95_0.pdf for a classification of simple abelian varieties over finite fields up to an isogeny.)

In the cases 2) and 3) $A$ becomes isogenous to a square of a supersingular elliptic curve over an algebraic closure of $k$.