jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then the equation $|z-3|=2|z+3|$ becomes $$|x+yi-3|=2|x+yi+3|,$$so $$(x-3{)}^2+y^2=4[(x+3{)}^2+y^2].$$This simplifies to $x^2+10x+y^2+9=0.$ Completing the square, we get $$(x+5{)}^2+y^2=4^2.$$This is the circle centered at $-5$ with radius 4.



The graph of $|z|=k$ is a circle centered at the origin with radius $k.$ We see that the circles with radius $\boxed{1}$ and $\boxed{9}$ intersect the circle $(x+5{)}^2+y^2=4^2$ in exactly one point.