= Solution
Write $C=V(P)$, where $P\in\mathbb C[x,y,z]$ is a <squarefree polynomial> and a <homogeneous polynomial> of degree $d$. A <projective point> $p\in C$ is a <singular point> precisely when
$$
P_x(p)=P_y(p)=P_z(p)=0.
$$
If $p=[p_0:p_1:p_2]$ is a <nonsingular point of an algebraic curve>, its <projective tangent line> is
$$
\boxed{T_pC=\{[x:y:z]:P_x(p)x+P_y(p)y+P_z(p)z=0\}.}
$$
The coefficients are not all zero, and the <Euler theorem for homogeneous functions> gives $p_0P_x(p)+p_1P_y(p)+p_2P_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the <projective point> only rescales the equation.
The strong form of <Bézout's theorem> says that two <projective plane curves> $C,D$ of degrees $d,e$, with no common <irreducible component>, have finitely many intersections and
$$
\boxed{\sum_{p\in C\cap D}I_p(C,D)=de.}
$$
Each <intersection multiplicity> is a positive integer. At $p\in C\cap D$,
$$
\boxed{I_p(C,D)=1\iff p\text{ is nonsingular on both curves and }T_pC\ne T_pD.}
$$
This is the criterion for a <transverse intersection of plane curves>. In particular, an intersection at a <singular point> of either curve contributes at least two.
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