The Kerr metric written using the Boyer-Lindquist coordinates is $$ds^2=-\frac{\Sigma\Delta}{A}dt^2+\frac{A\sin^2\theta}{\Sigma}(d\phi-\omega dt)^2+\frac{\Sigma}{\Delta}dr^2+\Sigma d\theta^2,$$ where $\Sigma=r^2+a^2\cos^2\theta$, $\Delta=r^2+a^2-2r$, $A=(r^2+a^2)^2-a^2\Delta\sin^2\theta$ and $\omega=2ar/A.$
Since the Kerr metric is not spherically symmetric, I am confused about the physical interpretation of the radial coordinate $r$.
My question: Is the radial coordinate of each point $(r,\theta,\phi)$ measured from the origin, as shown in the following image?
The Kerr metric has spheroidal symmetry, due to the presence of a non zero rotation. So Boyer-Lindquist coordinates are specially adapted for this symmetry. They are ‘oblate spheroidal coordinates’ but in disguise: if you look for them, you’ll see that this coordinates are given by $x=a \text{cosh} (\eta) \text{sin}(\theta) \text{cos}(\phi)$… and so on. If you call $r=a\text{sinh}(\eta)$, you obtain the Boyer-Lindquist coordinates expressed in terms of $(r,\theta,\phi)$. Then you see that the surfaces $r=\text{constant}$ are oblate spheroids (similar to a rugby ball). So no, $r$ is not a distance to the origin, it’s a coordinate that kind of indexes ‘what rugby ball you are on’. But note that, in the limit $a\rightarrow 0$, that is, when there is no rotation, you recover the usual spherical coordinates (and the Kerr line element turns to Schwarzschild line element).