Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either 
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching either circle.

Now, how I solved it was realising the line has a positive slope of $+2$, and thus if it's a tangent to circle #1 then its intercept should be negative. And so, using the condition of tangency,
$$a^2= m^2(r_1)^2+(r_1)^2= 4+1 $$
And thus, as $a$ can only be negative (otherwise a positive value of $a$ will make the line intersect the circle). Thus, making the lower bound of $a> - \sqrt 5$.
Similarly for the bigger circle $a<-\sqrt{20}=-\sqrt{(2^2)(2^2)+(2^2)}$
Hence I find that the solution should be $(-\sqrt5,-\sqrt{20})$, but the actual solution is $\left(2\sqrt 5-15,-\sqrt 5-1\right)$.

I would draw perpendicular to the line $2x - y + a = 0$ from centers of both circles.
From circle $C_1$ centered at $(1, 1)$,
$d_1 = \dfrac{|2 -1 + a|}{\sqrt5} \gt 1$
(where $1$ is the radius of the circle $C_1$).
Similarly from $(8, 1)$, $d_2 = \dfrac{|16 - 1 + a|}{\sqrt5} \gt 2$
(where $2$ is the radius of the circle $C_2$).
Simplifying we have,
$|a + 1| \gt \sqrt5$
$|a + 15| \gt 2 \sqrt5$
If the line is in between both circles, the algebraic perpendicular distances from centers will have opposite signs. In other words, $(a+1)$ and $(a+15)$ will have opposite signs.
i) From the first (considering both negative and positive value of $a+1$),
$a \gt \sqrt5 - 1 \ $ or $ \ a \lt - \sqrt5 - 1$
ii) From the second (again considering both negative and positive values of $a+15$),
$a \gt 2 \sqrt5 - 15 \ $ or $ \ a \lt - 2 \sqrt5 - 15$.
a) When $a \gt \sqrt5 - 1$, both $(a+1)$ and $(a+15)$ have positive signs.
b) When $a \lt - 2 \sqrt5 - 15$, both $(a+1)$ and $(a+15)$ have negative signs.
c) When $2 \sqrt5 - 15 \lt a \lt - \sqrt5 - 1$,  $(a+1)$ and $(a+15)$ have opposite signs.
That leads us to the answer $2 \sqrt5 - 15 \lt a \lt - \sqrt5 - 1$.
