In the 2D case each point (let's call it A) on the rigid body has two velocity components $(v_x,v_y)$ and the body itself rotates by $\omega$. The location of the ICR relative to A is $$icr = \begin{pmatrix} -\frac{v_y}{\omega} & \frac{v_x}{\omega} \end{pmatrix}$$
To prove this check that $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times \vec{r} = 0$ or with planar vectors $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} =
\begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} + \begin{pmatrix}0\\0\\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} $$
$$ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v_x - \omega \,y \\ v_y + \omega \,x \\ 0 \end{pmatrix} $$
$$ \begin{matrix} x=-\frac{v_y}{\omega} \\ y=\frac{v_x}{\omega} \end{matrix} $$
Also, since the body is rotating about the ICR the point A prescribes a circle (instantaneously) and thus yes the ICR is the center of curvature of all points on the rigid body.
In the 3D case a rigid body will rotate about an axis and it might have a parallel translation to that axis as well. Given a point A (on the body) with velocity $\vec{v}_A = (v_x,v_y,v_z)$ and rotational velocity of the body $\vec{\omega} = (\omega_x,\omega_y,\omega_z)$ the location if the ICR is given by
$$\vec{r}_{icr} = \frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} $$
In addition, the direction of rotation is $\vec{e} = \frac{\vec{\omega}}{\|\vec{\omega}\|}$ and the vector of any parallel translational motion is $$\vec{v}_\parallel = \left( \frac{\vec{\omega} \cdot \vec{v}_A}{\|\vec{\omega}\|^2}\right) \vec{\omega}$$
The scaling factor in the parenthesis above is called the screw pitch because the motion of rotation with parallel translation is a screw motion.
To prove this, look at the well known transformation law $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times \vec{r}_{icr} =0 $ and cross each side with $\vec{\omega}$
$$ \begin{array}{rcl}0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}_{icr}\right)\\0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\left(\vec{\omega}\cdot\vec{r}_{icr}\right)-\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)\\\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)&=&\vec{\omega}\times\vec{v}_{A}\\\vec{r}_{icr}&=&\frac{\vec{\omega}\times\vec{v}_{A}}{\vec{\omega}\cdot\vec{\omega}}=\frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} \end{array} $$
In step 3 above it is assumed that $\vec{\omega}\cdot \vec{r}_{icr} = 0$ which turns out to be true given that $\vec{r}_{icr}$ is perpendicular to the rotation axis.
