Of $n^2$ points of intersection, $np$ lie on curve of deg. $p < n$, then remaining $n(n - p)$ lie on a curve of deg. $n - p$

Question

Let $C$, $C'$ be two plane curves of degree $n$. Is the following statement true or not?

Suppose that of the $n^2$ points of intersection, $np$ lie on a curve of degree $p < n$, then the remaining $n(n - p)$ will lie on a curve of degree $n - p$.

Answer 1

Pick another point on the curve of degree $p$. Find a linear combination of the two curves of degree $n$ that vanishes at this new point. It intersects a curve of degree $p$ at $np + 1$ points, hence by the inverse of Bézout's theorem, they share an irreducible component. Assuming the curve of degree $p$ is irreducible, we then remove it from the degree $n$ curve to obtain a degree $n - p$ curve with the desired property.

If it is reducible, then we get that for each irreducible component, there is a linear combination containing both. If these are not all the same one, an intersection of both irreducible components of the degree $p$ curve is forced to be an intersection point of the two degree $n$ curves, hence counts as multiplicity greater than $1$ in Bézout's theorem, which contradicts there being $np$ intersection points.