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 $a+b$? 
 A: In a Euclidean plane, two circles centered at $O_1$ and $O_2$, with radii $r_1$ and $r_2$, respectively, intersect if and only if
$$\left|r_1-r_2\right| \leq O_1O_2 \leq r_1+r_2\,,$$
or equivalently,
$$\left|r_1-O_1O_2\right|\leq r_2 \leq r_1+O_1O_2\,.$$
Using Mohammad Zuhair Khan's hint, you would need $40+k\geq 0$ and
$$\left|7-\sqrt{40+k}\right|\leq \sqrt{(6-2)^2+(3-6)^2}\leq 7+\sqrt{40+k}\,.$$

 Thus, we have the equivalent condition $$2=|7-5|\leq \sqrt{40+k}\leq 7+5=12\,.$$  That is, $-36\leq k \leq 104$.  This should coincide with Dr. Sonnhard Graubner's approach.

A: 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:
$$2\le \sqrt{40+k}\le 12 \Rightarrow -36\le k\le 104.$$
A: Hint: Plugging $$y=\frac{8x+4-k}{6}$$ in to your equation we have to solve
$${k}^{2}-16\,xk+100\,{x}^{2}+28\,k-656\,x-272=0$$
Can you solve this?
A: let x^2+y^2-12x-6y-4=0 is A 
and let x^2+y^2-4x-12y-k=0 is B
considering A-B equals -8x+6y+k-4
and “A and B is equivalent to A and A-B” 
so we just consider k’s condition that A and A-B have simultaneous real solutions .
Ais a circle that center is (6,3) and radius is 7 
so A and A-B have simultaneous real solutions is circle A and line A-B have common point.
consider “the length between circle center and line A-B”
and the circle radius ,we can get the condition of k so get a+b.
