jmc

Completing the square in the first equation, we get $$(x-6{)}^2+(y-3{)}^2=7^2,$$which represents a circle centered at $(6,3)$ with radius 7.

Completing the square in the second equation, we get $$(x-2{)}^2+(y-6{)}^2=k+40,$$which represents a circle centered at $(2,6)$ with radius $\sqrt{k+40}.$



The distance between the centers is $\sqrt{4^2+3^2}=5,$ so the two circles intersect when the radius of the second circle is between $7-5=2$ and $7+5=12.$ This gives us $$2^2\le k+40\le 12^2,$$or $k\in \boxed{[-36,104]}.$