Composition of translation and rotation is a rotation, but what is its center? Consider the rotation $r_{\Omega,\alpha}$ of center $\Omega$ and angle $\alpha$.Furthermore let $t_{\vec{v}}$ be the translation by vector $\vec{v}$.
Then $$t_{\vec{v}}\circ r_{\Omega,\alpha}=r_{\Omega',\alpha}.$$
I am wondering how $\Omega'$ is defined. 

For $r_{\Omega,\alpha}\circ t_{\vec{v}}=r_{\Omega',\alpha}$ we have the relation $\Omega+\vec{v}=\Omega'$.
Is there a similar relation for $$t_{\vec{v}}\circ r_{\Omega,\alpha}=r_{\Omega',\alpha}.$$
Thanks in advance!
The formula you give is not correct, for instance, if you evaluate both side at $\Omega' = \Omega + \vec{v}$:

$$(r_{\Omega,\alpha} \ \circ \ t_{\vec{v}} \ )(\Omega + \vec{v}) = r_{\Omega,\alpha}(\Omega + 2\vec{v})\neq \Omega + \vec{v} = r_{\Omega',\alpha}(\Omega + \vec{v})$$

On the other hand there is a formula for the conjugation of a rotation with a traslation:

$$(r_{\Omega,\alpha})^{t_{\vec{v}}} := t_{\vec{v}} \ \circ \ r_{\Omega,\alpha} \ \circ \ t_{-\vec{v}} = r_{\Omega + \vec{v}, \ \alpha}$$
Translation commutes with rotation....