Periodic Sequence - version 2 - Radius of Trajectory

Let $\mathbf{v}(k) = \begin{bmatrix} x(k) \\ y(k) \\ z(k) \end{bmatrix}$

If

$\mathbf{v}(k+1) =A \mathbf{v}(k) + b$

where

$A =\begin{bmatrix} 0.944901543 &&-0.267677991 && 0.188439826 \\ 0.285722226 && 0.955319386&&-0.07568144 \\ -0.159761963&&0.125352956&&0.979164313\end{bmatrix}$

and $b = \begin{bmatrix} -0.266682736 \\ -0.420081571 \\ 0.3611953 \end{bmatrix}$

And $\mathbf{v}(0) = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}$

Then sequence $\{ \mathbf{v}(k) \}$ is a periodic sequence that lies on a circle in 3D space. Find the radius of this circle.