$x^2+y^2=4x+8y+5$ 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?
On
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.
On
My approach involves Calculus, which may reasonably be regarded as off-limits, given the euclidean-geometry and analytical-geometry tags.
$\color{red}{\text{However}}$:
See editing in $~\color{red}{\text{red}}~$ to see how this answer may be salvaged, without Calculus.
I agree that the locus of satisfying points $(x,y)$ is represented by the circle of radius $(5)$, centered at $(x,y) = (2,4)$. This is explicitly demonstrated by completing the square, so that the original equation is transformed into
$$(x^2 - 4x + 4) + (y^2 - 8y + 16) = 5 + 4 + 16.$$
Any line whose equation is $3x - 4y = m$ can have it's equation re-expressed as
$$y = \frac{3}{4}x - \frac{m}{4}.$$
Such a line has slope $\displaystyle \left(\frac{3}{4}\right).$
So, my plan of attack is:
Identify all points $(x,y)$ that satisfy both of the following equations:
$(x-2)^2 + (y-4)^2 = 5^2.$
$\displaystyle y' = \frac{dy}{dx} = \frac{3}{4}.$
Then, having found the two pertinent points $(x_1,y_1)$ and $(x_2,y_2)$ that satisfy both of the above requirements, I will compute the two values of $(m)$ such that the equations
$\displaystyle y_1 = \frac{3}{4}x_1 - \frac{m_1}{4}$
$\displaystyle y_2 = \frac{3}{4}x_2 - \frac{m_2}{4}$
are satisfied.
Then, having identified these two points as $m_1$ and $m_2$, I will infer that you must have that $m$ lies strictly between $m_1$ and $m_2$.
Using implicit differentiation,
$(x-2)^2 + (y-4)^2 = 25 \implies $
$2(x-2) + 2(y-4)y' = 0.$
Since $y'$ is required to be $\dfrac{3}{4}$, I am looking for the points on the circle such that
$$\frac{x-2}{4-y} = \frac{3}{4} \implies (12 - 3y) = (4x - 8) \implies $$
$$(20 - 4x) = 3y \implies y = \frac{20 - 4x}{3}. \tag1 $$
$\color{red}{\text{Salvaging the answer, without Calculus}}$
If the two pertinent points are $(x_1,y_1)$ and $(x_2,y_2)$, and if you construct the line segments
$$\overline{(2,4),(x_1,y_1)}, ~~~ \overline{(2,4),(x_2,y_2)}, $$
then (in the Analytical Geometry world), each of these line segments must be perpendicular to the required slope. This is because the line segment going from the center of the circle to the point of tangency must be perpendicular to the tangent line.
Therefore, the two line segments must each have slope $~\dfrac{-4}{3}.~$
That is, in $~\Bbb{R^2},~$ two (non-vertical, non-horizontal) lines are perpendicular, if and only if the product of their slopes is $(-1)$.
So, the two points $(x_1,y_1), (x_2,y_2)$ must each satisfy:
$$\frac{4 - y}{2 - x} = \frac{-4}{3},$$
which is equivalent to (1) above.
This implies that $x$ must satisfy
$$(x - 2)^2 + \left[\frac{20 - 4x}{3} - 4\right]^2 = 25 \implies $$
$$(x - 2)^2 + \left[\frac{8 - 4x}{3}\right]^2 = 25 \implies $$
$$(x - 2)^2 \left\{ ~\left[1\right]^2 + \left[\frac{-4}{3}\right]^2 ~\right\} = 25 \implies $$
$$(x - 2)^2 \left[1 + \frac{16}{9}\right] = 25 \implies $$
$$(x-2)^2 \times \frac{25}{9} = 25 \implies $$
$$(x - 2)^2 = 9 \implies (x - 2) \in \{\pm 3\} \implies $$
$$(y - 4) \in \{\pm 4\}.$$
Note
In the Analytical Geometry world, a shortcut is to realize that since the line segment going from the center of the circle to the two points of tangency has slope $~\dfrac{-4}{3}~$, you must have that the two triangles represented by
$$\triangle\{(2,4),(x,4),(x,y)\}$$
must each be a $3-4-5$ right triangle.
Therefore, the $4$ elements $\{(-1,0), (-1, 8), (5,0), (5,8)\}$ must be individually examined, to determine which of these elements satisfy (1) above, which is:
$$y = \frac{20 - 4x}{3}.$$
The pertinent points are $\{(5,0),(-1,8)\}.$
Therefore, the next step is to identify the two values $m_1, m_2$ such that
$$0 = \left[\frac{3}{4}\times (5)\right] - \frac{m_1}{4}, ~~ 8 = \left[\frac{3}{4}\times (-1)\right] - \frac{m_2}{4} \implies $$
$$m_1 = 15, m_2 = -35.$$
Therefore, you must have that
$$-35 < m < 15.$$
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.