Where does the line lie?

Let $C_{1}$ be $(x-1)^2 + (y-1)^2 = 1$ and $C_{2}$ be $(x-8)^2+(y-1)^2 = 4.$ If the line $y = 2x +a$ is to lie between these circles without touching/intersecting either one, then it shall include all such lines except the one tangent to $C_{1}$ in $x \in (1,2), y \in (0,1)$ and to $C_{2}$ in $x \in (6,8), y \in (1,3)$ . This also requires the necessary condition $a < 0.$ If we first examine $C_{1}$ , we obtain:

$x^2 + (2x+a)^2 - 2x - 2(2x+a) + 1 = 0 \Rightarrow 5x^2 +(4a-6)x + (a^2-2a +1) =0 \Rightarrow x = \frac{(6-4a) \pm \sqrt{(4a-6)^2 - 4(5)(a^2-2a+1)}}{2(5)} =\frac{(6-4a) \pm 2\sqrt{-a^2 -2a + 4}}{10}$

If the line is to be tangent to $C_{1}$ in exactly one point, then $-a^2-2a+4 = 0 \Rightarrow 5 = (a+1)^2 \Rightarrow a = -1-\sqrt{5} \approx -3.236.$

For $C_{2}$ , we have:

$x^2 + (2x+a)^2 - 16x - 2(2x+a) + 61 = 0 \Rightarrow 5x^2 +(4a-20)x + (a^2-2a +61) =0 \Rightarrow x = \frac{(20-4a) \pm \sqrt{(4a-20)^2 - 4(5)(a^2-2a+61)}}{2(5)} =\frac{(20-4a) \pm 2\sqrt{-a^2 -30a -205}}{10}$

If the line is to be tangent to $C_{2}$ in exactly one point, then $-a^2-30a - 205 = 0 \Rightarrow 20 = (a+15)^2 \Rightarrow a = -15+\sqrt{20} \approx -10.52.$

Hence, the solution set includes the lines $y = 2x + a$ with $\boxed{a \in (-10.52, -3.236)}.$