Projective cubic curve passing through seven points in $\mathbb{C}^2$ Let $x_1,...,x_7$ be distinct points in $\mathbb{C}^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.
My attempt so far... A related question was: show that for every five points $a_1,...,a_5\in\mathbb{P}^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.
Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.
 A: Let $\mathbb{P}^N\equiv\mathbb{P}(\mathbb{C}[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
\begin{equation}
\binom{3-1+3}{3}-1=10-1=9.
\end{equation}
Let $\displaystyle F=\sum_{i_0+i_1+i_2=3}a_{i_0i_1i_2}z_0^{i_0}z_1^{i_1}z_2^{i^2}$ be the generic homogeneous polynomial of degree $3$ in three variables; let
\begin{equation}
\nu_{2,3}:\mathbb{P}^2\to\mathbb{P}^9
\end{equation}
be the Veronese embedding of the (complex) projective plane in $\mathbb{P}^9$; one knows that the image of $F$ via $\nu_{2,3}$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $k\in\{1,\dots,7\}$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $\mathbb{P}^2$, constraint $H$ passing through $7$ distinct points of $\mathbb{P}^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $\mathbb{P}^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $\mathbb{P}^2$ gets through a unique cubic plane curve.
Let $F_{z_1}\equiv F_1$ and $F_{z_2}\equiv F_2$ be the partial derivaties of $F$; the curve $\Gamma=\{F(1:z_1:z_2)=0\}$ has a (unique) singular point if the system
\begin{equation}
\begin{cases}
F(1:z_1:z_2)=0\\
F_1(1:z_1:z_2)=0\\
F_2(1:z_1:z_2)=0
\end{cases}
\end{equation}
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $\Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $\Gamma$ is an elliptic curve.
