Proving non-centered at the origin ellipsoid A is a subset of ellipsoid B I would like to prove an ellipsoid $E_a$ center at $q_a$ with shape matrix $A$ and radius $r_a$ defined as:
$(x-q_a)^T A^{-1} (x-qa) <= r_a$
is included into another ellipsoid $E_b$ -i.e. $E_a$ covers $E_b$- defined as $E_a$:
$(x-q_b)^T B^{-1} (x-qb) <= r_b$
with x defined on $R^n$
I was thinking about showing that all points defined by $E_a$ are also included into $E_b$ such that $E_a <= E_b$ but I don't know how to prove a condition satisfying this ellipsoid status.
I don't really find paper or information about that ...
link
Because they target center at the origin ellipsoids.
Thank you a lot for your help and maybe if Jacob see this topic because he asked for the same question without any answers,
Regards.
 A: A possible approach to the problem is first to perform an affine transformation on $R^n$ that takes ellipsoid $E_b$ to the unit sphere.
Writing $u$ for the transformed coordinates, we have
$$u = \frac1{\sqrt{r_b}} B^{-1/2}(x - q_b).$$
Then $x = \sqrt{r_b} B^{1/2} u + q_b$ and
$$x - q_a = \sqrt{r_b} B^{1/2} u  + q_b - q_a 
= \sqrt{r_b} B^{1/2} \left( u - \frac1{\sqrt{r_b}} B^{-1/2}(q_a - q_b) \right).$$
Let
$q_0 = \dfrac1{\sqrt{r_b}} B^{-1/2}(q_a - q_b),$ and we can write the equation above more concisely as
$$x - q_a = \sqrt{r_b} B^{1/2} (u - q_0).$$
When we substitute the right-hand side of this equation for $x - q_a$ in the equation of ellipsoid $E_a,$ we get
$$(u - q_0)^T \left(\sqrt{r_b} B^{1/2}\right)^T A^{-1} \sqrt{r_b} B^{1/2} (u - q_0) \leq r_a,$$
which simplifies to
$$(u - q_0)^T B^{1/2} A^{-1} B^{1/2} (u - q_0) \leq \frac{r_a}{r_b}.$$
(Note that $B$ must be symmetric positive definite in order that
$(x-q_b)^T B^{-1} (x-q_b) \leq r_b$ may define an ellipsoid, and therefore
$(B^{1/2})^T = B^{1/2}.$)
Let $M = \dfrac{r_b}{r_a}B^{1/2} A^{-1} B^{1/2},$ and the problem reduces to a maximization problem in which we want to find 
$$s = \sup \{ u^T u : (u - q_0)^T M (u - q_0) \leq 1 \}.$$
Then $E_a$ is contained in $E_b$ if and only if $s \leq 1.$
You should be able to find helpful information on the solution of this maximization problem, for example on pages $13$-$14$ of 
this document.

As an aside, for an ellipsoid defined by $(x-q_b)^T B^{-1} (x-q_b) \leq r_b,$
I think it is misleading to call $r_b$ the "radius" of the ellipsoid.
First of all, an ellipsoid does not have a single radius;
furthermore, even if we do somehow single out one radius, for example the circumradius, you should find that as you vary $r_b$ in the equation above, the circumradius of the ellipsoid varies in proportion to $\sqrt{r_b},$ not $r_b.$
A more conventional way to write the equation of a general ellipsoid would be
$(x-q_b)^T B^{-1} (x-q_b) \leq 1,$
which you can get by scalar multiplication of the matrix you would have used in your original formula.
This would simplify the equations in this answer.
