Condition on $k$ for $x^2+y^2-12x-6y-4=0$ and $x^2+y^2-4x-12y-k=0$ to have simultaneous solutions $(x,y)$ The two equations: $$x^2+y^2-12x-6y-4=0$$ and $$x^2+y^2-4x-12y-k=0$$ 
have simultaneous real solutions $(x,y) \iff a \le k\le b$.
 Then, what is the value of $a+b$? 

The two equations represent the circles:
$$\begin{cases}x^2+y^2-12x-6y-4=0 \\ x^2+y^2-4x-12y-k=0 \end{cases} \Rightarrow \\
\begin{cases}(x-6)^2+(y-3)^2=7^2 \\ (x-2)^2+(y-6)^2=40+k \end{cases} \Rightarrow \\
\begin{cases}\text{center: A(6,3), radius: AC=AD=7} \\ \text{center: B(2,6), radius: min BC, max BD} \end{cases} $$
Refer to the graph:
$\hspace{1cm}$
Note that: $AB=\sqrt{(6-2)^2+(3-6)^2}=5$. Hence: $BC=2, BD=12$. It implies:
$$2\le \sqrt{40+k}\le 12 \Rightarrow -36\le k\le 104.$$
