Circle and straight line

As in the other solution, we find the equation of the circle to be $(x-10)^2+\left(y+\dfrac 9k\right)^2=27,$ so that we seek the range of $k$ for which there exist real $x$ satisfying $\pm \sqrt{27-(x-10)^2}-\dfrac 9k=y_L=\dfrac{-6}{k}x+6.$ Solving for $x$ gives $x=\dfrac{10k^2\pm 3k\sqrt{-k^2+68k-181}+36k+54}{k^2+36},$ which implies that $-k^2+68k-181\geq 0,$ and thus $34-5\sqrt{39}\leq k\leq 34+5\sqrt{39},$ making $a+b+c=\boxed{78}$ .

Completing the square, the circle equation can be written $(x - 10)^2 + (y + \tfrac 9 k)^2 = 27.$ The center of the circle is $C\:(10, -9/k)$ and the radius is $\sqrt{27}$ .

We rotate the entire coordinate system: $\begin{cases} x' = kx - 6y \\ y' = 6x + ky \end{cases}$ The rotation comes with a multiplication with factor $\sqrt{k^2 + 6^2}$ .

This transformation changes

• the line into the horizontal line $y' = 6k$ ;

• the center of the circle to $C'\:(10k + 54/k, 51)$ ;

• the radius of the circle to $\sqrt{27(k^2 + 6^2)}$ .

The circle intersects the line if the distance of the center of the circle to the horizontal line is less than the radius: $51 - 6k < \sqrt{27(k^2 + 6^2)} \\ 17 - 2k < \sqrt{3(k^2 + 6^2)} \\ (17 - 2k)^2 < 3(k^2 + 6^2) \\ k^2 - 68k + 181 < 0 \\ (k - 34)^2 < 975 \\ |k - 34| < 5\sqrt{39}.$ Thus $34 - 5\sqrt{39} \leq k \leq 34 + 5\sqrt{39}$ , so that $a + b + c = 34 + 5 + 39 = \boxed{78}.$