smc

Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0{)}^2+(y-y_0{)}^2=r^2$. The first curve becomes $(x-6{)}^2+(y-3{)}^2=7^2$, which is a circle centered at $(6,3)$ with radius $7$. The second curve becomes $(x-2{)}^2+(y-6{)}^2=40+k$, which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$. The distance between the two centers is $5$, and therefore the two circles intersect if $2\le r\le 12$. From $\sqrt{40+k}\ge 2$ we get that $k\ge -36$. From $\sqrt{40+k}\le 12$ we get $k\le 104$. Therefore $b-a=104-(-36)=\boxed{140}$.