Two circles touches each other externally at the point O. .. I am stuck on the following elementary problem that says:

Two circles touches each other externally at the point O. If PQ and RS are two diameters of these two circles respectively and $PQ || RS$ (Where || indicates parallel),then prove that P,O,S  are collinear. 


My Try: 
 I added the points P,O,S and also the points Q,O; ans R,O. 
From the given condition , $\angle POQ=90^{\circ}=\angle ROS$. [since,Any angle inscribed in a semi-circle is a right angle.]
Now,from the figure we see  $\angle POR=180^{\circ}-90^{\circ}=\angle QOS$.
And hence,$\angle POQ+\angle QOS=90^{\circ}+90^{\circ}=180^{\circ}=\angle POS$.
Hence ,we can conclude P,O,S are collinear.
Can someone verify it ? Am I right? 
Thanks in advance for your time. 

Flip the smaller circle (lets suppose its the one with diameter $RS$) about the common tangent at O, so that its now internally tangent. Let $R'$ and $S'$ be the reflected points corresponding to $R$ and $S$ resp.
The homothety centred at O now carries this circle into the larger circle, also carrying $R'$ to $P$ as well as $S'$ to $Q$ which means $OR'P$ is a straight line, and similarly $OS'Q$ is also a straight line.
So $ORS$ being a reflection of $OR'S'$ has $\angle POQ = \angle ROS$ which is what we need to conclude that $POS$ and $QOR$ are straight lines.
