When $k$ satisfies $a \leqslant k \leqslant b$, the graphs of $x^{2}+y^{2}-4-12 x+6 y$ and $x^{2}+y^{2}=k+4 x+12 y$ intersect, then $b-a$ equals
(A) 5
(B) 68
(C) 104
(D) 140
2. (D).
$x^{2}+y^{2}=4+12 x+6 y$, i.e., $(x-6)^{2}+(y-3)^{2}=7^{2}, x^{2}+y^{2}=k+4 x+12 y$, i.e., $(x-2)^{2}$ $+(y-6)^{2}=k+40$. They represent two circles, and the radius of the first circle is 7, the distance between the centers of the two circles is 5. To make the two circles intersect, it should be $2 \leqslant \sqrt{40+k} \leqslant 12$, solving this gives $-36 \leqslant k \leqslant 104$, so $b-a=140$.