The Cayley-Bacharach theorem (also known as the 9 point theorem or the $8 \rightarrow 9$ theorem states that if two cubic curves intersect in 9 points, and C is any curve through 8 of those points, then C also passes through the 9th point.
Suppose we replaced the word cubic with quartic or quintic? Suppose I have two curves of order $d$ that meet in $d^2$ points (by Bezouts), how many of these points does another degree $d$ curve C have to pass through before we an say it passes through all points?
My intuition tells me that the answer would be ${d+2} \choose 2$$ -2$ - but is this accurate? I derive this from the dimension of the vector space = number of linear conditions imposed by points + 2 (our two curves which we want to span the vector space of vanishing polynomials).
I am new to this area, so forgive any faux pas made with notation, convention, etc.
Cayley-Bacharach theorem indeed can be generalized to any plane curve. If $C_1$ and $C_2$ are plane curves, $\deg C_1 = n_1$ and $\deg C_2 = n_2$ meeting at $n_1 n_2$ distinct points, then any curve of degree $n_1+n_2-3$ that passes through all but one point of $C_1 \cap C_2$ also passes through the remaining point.
In this form it is proven in Griffith-Harris in chapter on residues, so you need to know multi-dimensional residues to understand the proof.
There is also very nice paper Cayley-Bacharach theorems and conjectures on the Cayley-Bacharach theorem, its applications and variations.