Bézout's theorem
Bézout's theorem for curves states that, in general, two algebraic curves of degrees $m$ and $n$ intersect in m·n points and cannot meet in more than $m·n$ points unless they have a component in common (i.e., the equations defining them have a common factor)
Question
Let $L = \left\{ {x + iy:x,y \in \mathbb{R},f(x,y) = \sqrt {{x^2} + {y^2}} p(x,y) + q(x,y) = 0} \right\}$ and $p,q$ are two polynomials. then $L$ is algebraic curve.
Let $(x,y)$ be singularity point,
By implict function theorem, we know that $\frac{{\partial {f_{}}}}{{\partial {x_{}}}}(x,y) = 0$ and $\frac{{\partial {f_{}}}}{{\partial {y_{}}}}(x,y) = 0$.
How can we prove, by Bézout's theorem, that $L$ has finitely many singularities?($x,y\in \mathbb{R}$)
First $f$ is not a polynomial so $f=0$ is not an algebraic curve. Without lose of generality, assume $p$ and $q$ are coprime to each other. (The general case is easy if you understand this coprime case). Show that $f=f_x=f_y$ has only finite many solutions. \begin{equation} \sqrt(x^2+y^2)p+q=0,\\ \frac{x}{\sqrt (x^2+y^2)}\,p+\sqrt(x^2+y^2)p_x+q_x=0\\ \frac{y}{\sqrt (x^2+y^2)}\,p+\sqrt(x^2+y^2)p_y+q_y=0 \end{equation} whose solutions is a subset of the solutions of the following algebraic equations \begin{equation} (x^2+y^2)p^2=q^2,\\ (x ~p+(x^2+y^2)p_x)^2=(x^2+y^2)q_x^2\\ (y~ p+(x^2+y^2)p_y)^2=(x^2+y^2)q_y^2\\ \end{equation} Use Bezout's theorem to show that the second set of algebraic equations only have finite solutions, then the solutions of the first non-algebraic equations would be a subset of a finite set, so it also has only finite solutions.
The basic idea is as above, but you need to take care of irreducibilities of equations to make the arguments sound, which I omit since it is not essential for the proof.