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 $(x,y)$

Question

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 $a+b$?

Answer 1

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?

Answer 2

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.

Answer 3

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.

Answer 4

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:

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Note that: $AB=\sqrt{(6-2)^2+(3-6)^2}=5$. Hence: $BC=2, BD=12$. It implies: $$2\le \sqrt{40+k}\le 12 \Rightarrow -36\le k\le 104.$$