$x^2+y^2=4x+8y+5$ 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?

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.
