$$ \begin{aligned} & x^2+y^2-16 x-20 y+164=r^2 \\ & \text { i.e. }(x-8)^2+(y-10)^2=r^2 \\ & \text { and }(x-4)^2+(y-7)^2=36 \end{aligned} $$ Both the circles intersect each other at two distinct points. Distance between centres $$ \begin{aligned} & =\sqrt{(8-4)^2+(10-7)^2}=5 \\ & \therefore|r-6|<5<|r+6| \\ & \therefore|f| r-6 \mid<5 \Rightarrow r \in(1,11) \\ & \text { and }|r+6|>5 \Rightarrow r \in(-\infty,-11) \cup(-1, \infty) \end{aligned} $$ From (3) and (4), $$ r \in(1,11) $$