The circle $$x^2 + y^2 = 4x + 8y + 5$$ intersects the line $$3x - 4y = m$$ at two distinct points if

The given circle is $$x^{2}+y^{2}=4x+8y+5$$ and the given line is $$3x-4y=m$$.

Step 1 : Write the circle in centre-radius form
Move all terms to the left and complete the squares:

$$x^{2}-4x+y^{2}-8y=5$$

Add and subtract the required constants:

$$(x^{2}-4x+4)+(y^{2}-8y+16)=5+4+16$$

$$\Rightarrow (x-2)^{2}+(y-4)^{2}=25$$

Hence, the centre is $$C(2,\,4)$$ and the radius is $$r=5$$.

Step 2 : Find the perpendicular distance of the centre from the line
For the line $$3x-4y-m=0$$, the perpendicular distance of $$C(2,4)$$ is

$$d=\frac{\lvert 3(2)-4(4)-m\rvert}{\sqrt{3^{2}+(-4)^{2}}} =\frac{\lvert 6-16-m\rvert}{5} =\frac{\lvert -10-m\rvert}{5} =\frac{\lvert m+10\rvert}{5}$$

Step 3 : Condition for two distinct points of intersection
The line cuts the circle in two distinct points when the distance $$d$$ is strictly less than the radius $$r$$:

$$\frac{\lvert m+10\rvert}{5}\lt 5 \;\;\Longrightarrow\;\; \lvert m+10\rvert \lt 25$$

This inequality expands to

$$-25 \lt m+10 \lt 25$$

$$\Rightarrow -35 \lt m \lt 15$$

Step 4 : Choose the correct option
The interval $$-35 \lt m \lt 15$$ matches Option A.

Option A which is: $$-35 < m < 15$$