Here is a sketch of a way to so this.
Let the points be $P, Q$ on circles $A, B$ with centres $O_A, O_B$, and $a,b$ be the points where the extension of the segment joining the centres meets the two circles.
Then we have $$PQ\le PO_A+O_AQ=aO_A+O_AQ\le aO_A+O_AO_B+O_BQ=aO_A+O_AO_B+O_Bb=ab$$
This uses "any two sides of a triangle are together greater than the third" for the inequalities (which may be equalities if the "triangle" is degenerate), and the fact that two radii of the same circle are equal for the first two equalities. Worth drawing a diagram.
Then you have to use the information you have about the circles to calculate the length in your specific case.