Suppose we wish to observe a Euclidean circle $C$ with radius $\alpha$. We define the relation $$R=\{(x,y)\in\mathbb{R}^2:\alpha^2=x^2+y^2,\text{fixed}\,\alpha\in\mathbb{R}^{+}\},$$ represented in Cartesian $(x,y)$ coordinates. By means of calculus, we manipulate the derivative $\frac{\text{d}y}{\text{d}x}$ and obtain the following one-parameter family of equations for the orthogonal trajectories to $C$: $$y=\kappa x.$$ Simple enough, though I suspect a caveat. If we say $\kappa\in\mathbb{R}$, then the vertical line given by $x=0$ is not included in our family of orthogonal curves. Likewise, nothing stops us from totally differentiating with respect to $y$ and obtaining $x=\kappa y$, which fails to include $y=0$. Is this a cause for concern? Or are we to implicitly consider the limiting case as well?
I believe alternative methods on the polar equation permit $\theta$ to take on any real value, which seems more complete than the above results. And there are several examples of other curves whose orthogonals have such a peculiarity. So what's going on here? Are my math or interpretation wrong (quite possibly), or is there truly a limitation with the classical method in Cartesian coordinates?
P.S. I'm a first-year undergrad. I understand that higher math (i.e. complex analysis and perhaps topology) has the sufficient tools to approach the subject more strongly, so for sake of completeness I don't mind mathematically mature answers that are well beyond me.
Your problem stems from treating $x$ and $y$ differently.
Your curve $C$ is a level line of the function $$f(x,y):=x^2+y^2\ .$$ It is a basic fact of multivariate calculus that at any point ${\bf z}_0=(x_0,y_0)\in C$ the gradient $\nabla f({\bf z}_0)$ is orthogonal to $C$ at ${\bf z}_0$. Therefore the normal line $n$ to the curve at ${\bf z}_0$ is given by $$n:\quad t\mapsto {\bf n}(t):={\bf z}_0+t\>\nabla f({\bf z}_0)\qquad(-\infty<t<\infty)\ ,$$ or written out in coordinates: $$n:\quad t\mapsto \left\{\eqalign{ x(t)&:=x_0+t\>f_x(x_0,y_0)\cr y(t)&:=y_0+t\>f_y(x_0,y_0)\cr}\right.\ .\tag{1}$$ In the example at hand we have $f_x(x_0,y_0)=2x_0$, $f_y(x_0,y_0)=2y_0$. If you plug this into $(1)$ you get $$n:\quad t\mapsto \left\{\eqalign{ x(t)&:=(1+2t)x_0\cr y(t)&:=(1+2t)y_0\cr}\right.\qquad(-\infty<t<\infty)\ .\tag{2}$$ Here $t=0$ corresponds to the "origin" ${\bf z}_0$ of $n$. If you eliminate $t$ from $(2)$ you obtain the equation $$n=\bigl\{(x,y)\>\bigm|\>y_0 x-x_0y=0\bigr\}$$ for $n$ – without certain exceptional points ${\bf z}_0\in C$.