Let there be two circles $C(A, R_A)$ and $C(B, R_B)$, one with center $A$ and radius $R_A$, the other with center $B$ and radius $R_B$. Let $P$ and $Q$ be the farthest points of the two circles, as on the diagram below.
Draw the tangents from $P$ to $C(B, R_B)$ and from $Q$ to $C(A, R_A)$. Whenever the construction is possible, it leads to two "isosceles triangles" with one side a circular arc.
The fact is that the "incircles" of the two triangles are always equal, i.e., have the same radius.
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