$x^2+y^2=4x+8y+5$ intersects $3x-4y=m$ at two distinct points 
$x^2+y^2=4x+8y+5$ intersects $3x-4y=m$ at two distinct points. What are $m$'s possible values?

I got that the centre of the circle is $C=(2,4)$ and its radius is $r=5$. If the line $l$ intersects the circle at two points, the distance between the line and $C$ is less than $r$:
$$\frac{|6-16-m|}5<5\implies-35<m<15$$
Is this correct?
 A: Your solution looks fine. If this was on a test or an assignment or something that I was correcting, I would personally prefer it if you added an intermediate step or two on how you found the circle, on how you set up the inequality, and on how you solved it. But that's a matter of presentation, not of math.
Alternately, consider the function in the plane given by $$x^2+y^2-4x-8y-5$$ It is zero exactly on the circle. Now restrict this function to the line. On the line, the value of $y$ is the same as the value of $\frac{3x-m}4$. So on the line, the entire function has the same value as $$x^2+\left(\frac{3x-m}4\right)^2-4x-8\cdot\frac{3x-m}4-5$$Whether this has zero, one, or two roots is determined by the discriminant of that quadratic. Which gives you a quadratic inequality to solve.
Is this more work? Looks like it is. But this way you can solve similar problems with ellipses and hyperbolas with little to no modification, which isn't true for your approach. Does that make one approach better than the other? Not at all. They are good at different things.
A: An alternative approach is, solve $3x - 4y = m$ for $y$.  Plug that into the other equation:
$$x^2 + \left(\frac{3x-m}{4}\right)^2 = 4x + 8\frac{3x-m}{4} + 5$$
Then solve for $x$ using quadratic:
$$x=\frac{3m+80 \pm 4\sqrt{-m^2-20m+525}}{25}$$
2 solutions will exist when $-m^2-20m+525 = -(m-15)(m+35) > 0$
This will tell you whether your final result is correct.
