<s>[INST] 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 $\left(2\sqrt 5-15,-\sqrt 5-1\right)$.
 [/INST] The line passing through the center of the first circle is $y=2x-1$. Its perpendicular line passing through its center is $y=-\frac12x+\frac12$. It crosses the first circle at $(x,y)=\left(1+\frac2{\sqrt5},1-\frac1{\sqrt5}\right)$. Hence, $y=2x-\sqrt5-1$ touches the first circle from the right side.
Similarly, the line passing through the center of the second circle is $y=2x-15$. Its perpendicular line passing through its center is $y=-\frac12x+5$. It crosses the first circle at $(x,y)=\left(8-\frac4{\sqrt5},1+\frac2{\sqrt5}\right)$. Hence, $y=2x+2\sqrt5-15$ touches the second circle from the left side.
Hence, the value of $a$ ranges between $2\sqrt5-15$ and $-\sqrt5-1$.
See the Desmos graph for reference.
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