Four circles (no two congruent) have centers $A,B,C,D$, with $P,Q$ on all four. Circle $A$'s radius is $\frac59$ circle $B$'s, and circle $C$'s is $\frac59$ circle $D$'s. Also $AB=CD=56$ and $PQ=54$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$?

1. With $AR=x,BR=y$ satisfying $y-x=56$ and $x^2=a^2-27^2$, $y^2=\left(\frac95\right)^2a^2-27^2$: solving gives $x=61,y=117$.
2. As before, the only realizable configuration for $C,D$ has $CR+DR=CD=56$.
3. $AR+BR+CR+DR=61+117+56=\boxed{234}$.