Given a Quadratic Surface in the form:
$ax^2+by^2+cz^2+2fyz+2gzx+2hxy+2px+2qy+2rz+d=0$
I know how to decide which kind of surface is represented (http://mathworld.wolfram.com/QuadraticSurface.html). How can I find center and axis of this Quadratic Surface?
the Hessian matrix $H$ of second partials picks out the quadratic form terms. We can name a column vector $p$ and write your polynomial as $$ \frac{1}{2} x' H x + p' x + d, $$ where $x', p'$ are the transposes of the column vectors $x,p.$
If $H$ is invertible there is a center. Written as a column vector, the gradient of the polynomial is $$ H x + p. $$ The center is at $- H^{-1}p.$ Indeed, if we write $$ x = y - H^{-1} p, $$ your polynomial comes out to $$ \frac{1}{2} y' H y + \left( d - \frac{1}{2} p' H^{-1} p \right) $$ At this point you still have an eigenvector problem to find an axis.
If $H$ is not invertible various things may happen...