Since $\angle POA = \angle POB = 1$ right angle, $\angle AOB$ at the point of contact $O$ is a—

With each radius perpendicular to the common tangent, $\angle POA$ and $\angle POB$ are both right angles lying on opposite sides of the tangent. Together they form $90^{\circ} + 90^{\circ} = 180^{\circ}$, so is a straight angle, confirming the centres and contact point are collinear.