TITLE: Can a curve intersect a given curve only at given points?
QUESTION [15 upvotes]: Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$? 
I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case:

If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem).
If $X$ is a rational curve and $X \setminus \{P\} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = \{P\}$.
Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = \{P\}$ iff $P$ is a torsion point in the group.

If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable.
 Edit:  I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points.

REPLY [2 votes]: This is actually an answer (a bit belated!) to one of the questions Juan posed in his answer. I did not want to put it as just a comment, since I thought the question was good, and the answer is cute enough to be interesting to some people. 
The answer to Juan's first question is affirmative: "Every finite set of points in the projective plane is a set-theoretic complete intersection." Here are the arguments: we may assume number of points is $d + 2$, $d \geq 0$. Choose $\mathbb{C}^2 \subseteq \mathbb{P}^2$ and coordinates $(x,y)$ on $\mathbb{C}^2$ such that 

one of the points is on the intersection of $y$-axis and the line at infinity, 
The others are in $\mathbb{C}^2$,
The finite points have mutually distinct $x$-coordinates.

Now by Lagrange interpolation we can find a polynomial curve $C$ with equation of the form $y = f(x)$ which passes through each of the finite points. We may (and will) assume $\deg(f) \geq 2$, so that $C$ passes also through the other point which is at infinity. Let $a_1, \ldots, a_{d+1}$ be the $x$-coordinates of the finite points. Let $g_1 := z\prod_{i=1}^{d+1}(x - a_1z)$ and $g_2(x,y,z)$ be the homogenization of $y - f(x)$ (with respect to $z$). Then $g_1 = 0$ and $g_2 = 0$ intersects precisely at the given points - answering Juan's question.
Note that $g_2$ is irreducible. Replacing $g_1$ by an element of the form $g_1^p + \lambda g_2^q$ for suitable $\lambda$, $p$ and $q$, we may ensure that $g_1$ is also irreducible. Is it possible to ensure that the curves are non-singular?