$26 \cdot 29$ The graphs of the functions $x^{2}+y^{2}=4+12 x+6 y$ and $x^{2}+y^{2}=k+4 x+12 y$, for different values of $k$ that satisfy $a \leqslant k \leqslant b$, always intersect, then the value of $b-a$ is
(A) 5 .
(B) 68 .
(C) 104 .
(D) 140 .
(E) 144 .
(49th American High School Mathematics Examination, 1998)
［Solution］The given two functions can be transformed into:
$$
\begin{array}{l}
(x-6)^{2}+(y-3)^{2}=7^{2} \\
(x-2)^{2}+(y-6)^{2}=40+k
\end{array}
$$

They represent two circles with centers at $(6,3)$ and $(2,6)$, and radii of 7 and $\sqrt{40+k}$, respectively.

The two circles intersect $\Leftrightarrow R_{1}-R_{2} \leqslant$ distance between the centers $\leqslant R_{1}+R_{2}$, where $R_{1}$ and $R_{2}$ are the radii of the two circles.
Thus, we have
$|7-\sqrt{40+k}| \leqslant \sqrt{(6-2)^{2}+(3-6)^{2}} \leqslant 7+\sqrt{40+k}$.
That is, $|7-\sqrt{40+k}| \leqslant 5 \leqslant 7+\sqrt{40+k}$,
The right inequality always holds, so we only need to solve $|7-\sqrt{40+k}| \leqslant 5$.
This gives $2 \leqslant \sqrt{40+k} \leqslant 12$, leading to $-36 \leqslant k \leqslant 104$.
$$
\therefore b-a=140 \text {. }
$$

Therefore, the answer is $(D)$.