How can I define the direction of the normal form of concentric circle? $x(t)=C_1\left( \begin{array}{ccc} \cos\beta t\\ -\sin\beta t\\ \end{array} \right) +C_2\left( \begin{array}{ccc} \sin\beta t\\ \cos\beta t\\ \end{array} \right)$
I have no idea how to define the circle direction of concentric circle
I am not fully sure, but maybe this helps. You have $x(t) = \begin{pmatrix} \cos(\beta t) & \sin(\beta t)\\ -\sin(\beta t) & \cos(\beta t) \end{pmatrix} x_0$, which is expressing a clockwise rotation for the usual $(\beta, t)$ (changed the $C$'s to $x_0$). Then by differentiation find that $\dot{x}(t) = \begin{pmatrix} -\sin(\beta t) & \cos(\beta t)\\ -\cos(\beta t) & -\sin(\beta t) \end{pmatrix} \beta x_0$ such that for example when given $x_0=(1,0)$ the corresponding derivative (direction) becomes $\dot{x}(t)|_{t=0}=(0,-\beta)$. Which, if you draw a radius $1$ circle with $x_0$ on it, is what you should expect.