E.g. Looking back at some of previous work, I can't understand how I solved this question in the way that I did:
I would now solve this using the fact that $m_{radius} = \frac{1}{3}$ and the centre $=(5, -3)$ with $y-y_1 = m(x-x_1)$ to find the equation of the radius and diameter to such tangents. I would then solve this equation along with the equation of the circle as a system of equation to find the points where l intersects with C. Finally, I would then sub into $y-y_1 = m(x-x_1)$ to obtain the possible equations for l.
However, that's not how I actually solved this problem. Instead, I solved this using $m_{radius} = \frac{1}{3}$, centre $=(5, -3)$, and that $m = \frac{y_2-y_1}{x_2-x_1}$ for a straight line. Picking the centre as $(x_1, y_1)$, I solved $\frac{1}{3}= \frac{y_2+3}{x_2-5}$ as if the numerator and denominator were each individual equations:
$ y_2 + 3 = 1$
$y_2 = -2$
$ x_2 - 5 = 3$
$x_2 = 8$
I then repeated this process for the other point by picking the centre as $(x_2, y_2)$ and solving for $(x_1, y_1)$. Thus, I ended up with $(8, -2)$ and $(2, -4)$. I subbed this into $y-y_1 = m(x-x_1)$ to obtain the correct possible equations for l, $y = -3x + 22$ and $y = -3x + 2$.
What I don't understand is why solving $\frac{1}{3}= \frac{y_2+3}{x_2-5}$ as if the numerator and denominator were individual equations worked to find the points I needed. Surely $\frac{1}{3}= \frac{y_2+3}{x_2-5}$ is underdetermined as there are two unknowns? Is there some property of it being ratio with one unknown in the numerator and one unknown in denominator that means this is not an issue? Are there are other instances where this is the case? Moreover, I believe $\frac{1}{3}= \frac{y_2+3}{x_2-5}$ has to be underdetermined. Surely, if the centre is $(x_2, y_2)$, then $(x_1, y_1)$ could be any point before $(5, -3)$ along the line with the gradient $m = \frac{1}{3} $? Likewise, $(x_2, y_2)$ could be any point after $(5, -3)$ along this line.
With the 'correct' method, it's only with the additional constraint $(x-5)^2 + (y+3)^2 = 10$ that the points we are trying to find are narrowed down to $(8, -2)$ and $(2, -4)$. So, I can't understand how I could solve this problem without that constraint.
This is a fluke. If you have an equation that looks like $$\frac ab = \frac xy$$ and you are able to solve it by solving $a=x$ and $b=y$ separately, then you are a very lucky person. If you define $c=2x$ and $d=2y$, then it's still true that $$\frac ab = \frac cd$$ Then we do not have $a=c$ (unless $a=0$) and we definitely do not have $b=d$.
The problem was designed to have "nice" numbers, as is usual for problems that do not arise naturally, and it turned out that the ratios worked out just right for you to be able to solve it this way. But there's no reason for it to work, because you can multiply both the top and the bottom by any constant factor to get a wrong equation in the same way.