The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if
The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if
$-35 < m < 15$
$15 < m < 65$
$35 < m < 85$
$-85 < m < -35$
Solution
Circle $x^2+y^2-4 x-8 y-5=0$
Centre $=(2,4)$, Radius $=\sqrt{4+16+5}=5$
If circle is intersecting line $3 x-4 y=m$
at two distinct points.
$\Rightarrow$ length of perpendicular from centre $ < $ radius
$\Rightarrow \frac{|6-16-\mathrm{m}|}{5} < 5$
$\Rightarrow|10+\mathrm{m}| < 25$
$\Rightarrow-25 < m+10 < 25$
$\Rightarrow-35 < \mathrm{m} < 15$.