$L$ is algebraic curve. Why has $L$ finitely many singularities?($x,y\in \mathbb{R}$)

Question

Let

${L_1} = \left\{ {x + iy:x,y \in \mathbb{R},{f_1}(x,y) = (\sqrt {{x^2} + {y^2}} )p{{(x,y)}^{}} + q{{(x,y)}^{}} = 0} \right\} \subseteq {L_2} = \left\{ {x + iy:x,y \in \mathbb{R},{f_2}(x,y) = ({x^2} + {y^2})p{{(x,y)}^2} - q{{(x,y)}^2} = 0} \right\}$

and $p,q$ are two polynomials.Assume that $p$ and $q$ are coprime ($L_2$ is algeraic curve.)

Why has $L_1$ finitely many singularities?(How can we prove, by Bezout's theorem )

Answer 1

The singular points of the curve with equation $f=0$ are the solutions of the system of equations $$f=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0.$$ In other words, the set of solutions is the intersection of the three curves \begin{align} &f=0, && \frac{\partial f}{\partial x}=0, && \frac{\partial f}{\partial y}=0. \end{align}

Can you use Bézout's theorem to bound the number of points in this intersection?