External Center of Similitude of Circles with respect to Radii

Theorem
Let $A$ and $B$ be the centers of two circles $\bigcirc Ar$ and $\bigcirc BR$ whose radii are $r$ and $R$ respectively, $r \ne R$.

Let $\bigcirc Ar$ and $\bigcirc BR$ be such that neither is completely enclosed inside the other.

Let $T$ be the external center of similitude of $\bigcirc Ar$ and $\bigcirc BR$.

Let $P$ and $Q$ be points on the circumference of $\bigcirc Ar$ and $\bigcirc BR$ respectively.

Then:
 * the radii $AP$ and $QR$ are parallel


 * $T$, $P$ and $Q$ are collinear.
 * $T$, $P$ and $Q$ are collinear.

Proof

 * External-Center-of-Similitude.png

Sufficient Condition
Let $AP$ and $QR$ are parallel.

We note that:
 * $\angle TAP = \angle TBQ$
 * $\angle PTA = \angle QTB$

and so by Triangles with Two Equal Angles are Similar:
 * $\triangle TAP$ and $\triangle TBQ$ are similar.

Hence:
 * $\dfrac {AT} {BT} = \dfrac r R$

As $P$ and $Q$ are arbitrary, it follows that the straight lines connecting the ends of all such parallel radii pass through $T$.

The same applies to the tangents, as they touch $\bigcirc Ar$ and $\bigcirc BR$ at parallel radii.

Necessary Condition
Let $T$, $P$ and $Q$ be collinear.

Let $TP$ and $TQ$ cut $\bigcirc Ar$ and $\bigcirc BR$ at $P'$ and $Q'$.

Then $AP$ and $BQ$ are parallel as are $AP'$ and $BQ'$.