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. 
 A: 
One result very known is that: both centers and the tangent point are collinear. So $A,O,B$ are collinear.
One way to prove that $P,O,S$ are collinear is conclude that $\angle AOP=\angle BOS$.
Once $\overline{PQ} \parallel \overline{RS}$ then $\angle PAO=\angle OBS$ (Thales Theorem) and also:
$$\frac{AP}{AO}=\frac{OB}{BS}=1$$
then the triangles $PAO$ and $OBS$ are similars by the case SAS. That give us $\angle AOP=\angle BOS$ and then $POS$ are collinear.
A: I do not think that your approach is valid, since you have assumed that $QOR$ and $POS$ are collinear and then calculated
$\angle POR=180^{\circ}-90^{\circ}=\angle QOS$
What you can do instead, is to draw the line between centers of circles. The center of the left circle is $C_1$ and the right one is $C_2$. Also, draw a line from $P$ to $S$. Note that we do not know if $PS$ passes $O$. The line that passes through the centers meets $PS$ at $X$. Now, you can easily see that triangles $PC_1X$ and $SC_2X$ are similar. Therefore, the ratio of their edges is
$\frac{|C_1P|}{|C_2S|}=\frac{|C_1X|}{|C_2X|}=\frac{r_1}{r_2}$
Where $r_1$ and $r_2$ are radius of the circles.
Therefore, we know that $X$ is on the line $C_1C_2$ and it also has to satisfy the ratio. So, it should be the same as $O$.
To be more specific, we have two equations and two unknowns
$|C_1X|+|C_2X|=r_1+r_2$
$\frac{|C_1X|}{|C_2X|}=\frac{r_1}{r_2}$
A: 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.
A: $ POQ,ROS$ are right angles because they are contained in a semi-circle, by Thales theorem.
EDIT1:
My bad, an error occured before. 
Imagine two disjunct  right triangles hinged at right angle vertex with discontinuous slope. Unless alternate angles at diameters are equal a same line continuity  cannot be established. When cyan right triangle at right is rotated so that $ \beta = B $ these are alternate angles. So the transversal cutting line through $O$ has to be single continuous cutting line.

