help!!!

Solving the quadratic equations, we get -40 <= k <= 12.

it was wrong D:

Find all values of k so that the graphs of x^2+y^2=4+12x+6y and x^2+y^2=k+4x+12y intersect. Enter your answer using interval notation.

\(x^2+y^2=4+12x+6y\;\; and \;\;x^2+y^2=k+4x+12y \\~\\ x^2-12x+y^2-6y=4\\ x^2-12x+36+y^2-6y+9=4+36+9\\ (x-6)^2+(y-3)^2=49\\ \text{circle centre (6,3) radius 7}\\~\\ x^2+y^2=k+4x+12y\\ x^2-4x+y^2-12y=k\\ (x-2)^2+(y-6)^2=k+40\\ \text{circle centre (2,6) radius} =\sqrt{40+k} \)

Now find the distance between the 2 centres.

I got 5

5 is less than 7 so the centre of the variable circle is inside the fixed circle

It the radius of the variable circle is less than 2 then it will be entirely inside the other one and there will be no intersection.

I leave it to you to check everything I have done very carefully and to finish it off neatly.

Please no one provide any more details but asker guest can ask for clarification on details if they want to.

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Coding only:

x^2+y^2=4+12x+6y\;\; and \;\;x^2+y^2=k+4x+12y \\~\\

x^2-12x+y^2-6y=4\\
x^2-12x+36+y^2-6y+9=4+36+9\\
(x-6)^2+(y-3)^2=49\\
\text{circle centre (6,3) radius 7}\\~\\
x^2+y^2=k+4x+12y\\
x^2-4x+y^2-12y=k\\
(x-2)^2+(y-6)^2=k+40\\
\text{circle centre (2,6)  radius} =\sqrt{40+k}