(2) As shown in the figure, in the dihedral angle $\alpha-l-\beta$ of $120^{\circ}$, circle $O_{1}$ and circle $O_{2}$ are in the half-planes $\alpha$ and $\beta$ respectively, and are tangent to the edge $l$ at the same point $P$, then the sphere with circle $O_{1}$ and circle $O_{2}$ as cross-sections ( ).
(A) has only 1
(B) has only 2
(C) has infinitely many
(D) does not exist
(2) Connect $P O_{1}$ and $P O_{2}$, then $P O_{1} \perp l, P O_{2} \perp l$, so $\angle O_{1} P O_{2}$ is the plane angle of the dihedral angle $\alpha-l-\beta$. In the plane $O_{1} P O_{2}$, draw perpendiculars to $P O_{1}$ and $P O_{2}$ through points $O_{1}$ and $O_{2}$ respectively, and let the intersection of these two perpendiculars be $O$. Since $\angle O_{1} P O_{2}=120^{\circ}$, such an intersection point $O$ is unique. Therefore, there is exactly one sphere with the circles $O_{1}$ and $O_{2}$ as cross-sections, with the center of the sphere being $O$ and the radius being $O P$. Hence, the answer is A.