Denote $A(2, 4)$, $B(3, 3)$. Then circle $\Omega$ passes through points $O$, $A$ and $B$. Note that $\angle OBA = 90^\circ$ (the slopes of lines $OB$ and $AB$ are $1$ and $-1$, respectively), so $OA$ is a diameter of circle $\Omega$. Consequently, the maximum of the distance from a point on circle $\Omega$ to the origin $O$ is $|OA| = 2\sqrt{5}$.