Surface described by ${\bf r^\top A r + b^\top r}=1$? I was asked to describe the surface described by
$${\bf r}^\top {\bf A} {\bf r} + {\bf b}^\top {\bf r} = 1,$$
where $3 \times 3$ positive definite matrix ${\bf A}$ and vector $\bf b$ are given.
My intuition tells me that it is a rotated ellipsoid with a centre that is off the origin. However, I am told to show this via the substitution ${\bf r} = {\bf x} + {\bf a}$, with $\bf a$ being a constant vector, and dictate the conditions on this vector to obtain a new quadric surface ${\bf x}^\top{\bf A}{\bf x} = C$. However, upon substitution, I get a ridiculously messy answer involving combinations of position and constant vectors. Is there a trick I am missing out on? Thank you!

Note: for convenience we use $2b$ instead of $b$.
$$(x+a)^TA(x+a)+2b^T(x+a)=x^TAx+a^TAx+x^TAa+a^TAa+b^Tx+2b^Ta.$$
Notice that by symmetry of $A$, $x^TAa=a^TAx$. Collecting all the $x$ terms,
$$(2a^TA+2b^T)x$$ can be cancelled with the choice
$$a=-A^{-1}b.$$
Then
$$C=1-a^TAa-2b^Ta=1+b^TA^{-1}b>1.$$
(As $A$ is positive definite, it is invertible. Note that this is a matrix version of the "complete the square" paradigm.)

Going further, you can diagonalize the matrix and write, in the basis defined by the Eigenvectors
$$y^T\Lambda y=C$$
or
$$\lambda_0u^2+\lambda_1v^2+\lambda_2w_2=C.$$
As the three Eigenvalues are positive,
$$(\sqrt\lambda_0u)^2+(\sqrt\lambda_1v)^2+(\sqrt\lambda_2w)^2=C$$ describes a stretched sphere.
