I was to prove the following proposition from an old Tournament of Towns problems archive:
Problem. A circle $\omega_{1}$ with center $O_{1}$ passes through the center $O_{2}$ of another circle $\omega_{2}$. The tangent lines to $\omega_{2}$ from a point $C$ on $\omega_{1}$ intersect $\omega_{1}$ again at points $A$ and $B$. Prove that $AB$ is perpendicular to $O_{1}O_{2}$.
The proof I have come up with (below) seems slightly weird to me, in that I construct the second circle $\omega_{2}$ from an arbitrary inscribed triangle $ABC$ in the first circle $\omega_{2}$ with perpendicular $O_{1}O_{2}$ to $AB$, instead of directly proving that $AB$ is perpendicular to $O_{1}O_{2}$. I think this is right since everything I started with is arbitrary, but it still seems like I may be missing something. I wanted to know if there are any holes to my logic (or non-logic).
Proof. Let $ABC$ be an inscribed triangle of $\omega_{1}$, and let $O_{1}O_{2}$ be the radius of $\omega_{1}$ that is a perpendicular bisector of $AB$. Then arcs $AO_{2}$ and $BO_{2}$ are equivalent in measure, whence $\angle ACO_{2}$ and $\angle BCO_{2}$ are of the same degree measure. Let $O_{2}P$ and $O_{2}Q$ be perpendiculars to the extended lines $CA$ and $CB$, respectively. Then it is now immediate that triangles $PCO_{2}$ and $QCO_{2}$ are similar by their angles. In fact, they are congruent since they share side $CO_{2}$. Thus, $|AO_{2}| = |BO_{2}|$, and these are radii of a circle $\omega_{2}$ with center $O_{2}$ that has the extended lines $CA$ and $CB$ as tangents.
Since there is a unique circle $\omega_{2}$ with center $O_{2}$ as described above, I think the logic in the proof works.
Your "reverse" reasoning looks correct, but I think a more direct approach would be easier. As $CA$ and $CB$ are tangent to $\omega_2$ then $CO_2$ is the bisector of $\angle ACB$. It follows that arcs $AO_2$ and $BO_2$ are congruent and so are their chords: $AO_2\cong BO_2$. But then triangles $AO_1O_2$ and $BO_1O_2$ are congruent and it follows immediately that $AB\perp O_1O_2$.