Thank you Bill...
Indeed after running the command you suggested, PolarPlot[] likes %[[2]].
PolarPlot[]
%[[2]]
Perhaps
Solve[ReplaceAll[y==3 x+2, {x :> r Cos[?], y :> r Sin[?]}], r][[1, 1]] /. Rule[p_, q_] -> Equal[p, q]
But I am not certain that is always going to provide the result in the form you want.
