minimum exists because ${\cal C}$ is a closed set.

\[D-distance1\] Let ${\cal C} $ be an algebraic plane curve over ${\Bbb C}$ with an infinity branch $B$. We say that a curve ${\overline{{\cal C}}}$ [*approaches*]{} ${\cal C}$ at its infinity branch $B$ if there exists one leaf $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb
C},\,|z|>M\}\subset B$ such that $\lim_{z\rightarrow\infty}d((z,r(z)),\overline{\cal C})=0.$

We will show that this condition is satisfied for one leaf of $B$ if and only if it is satisfied for every leaf of $B$. It will be derived as a consequence of the following theorem.

\[T-curvas-aprox\] Let ${\cal C}$ be a plane algebraic curve over $\Bbb C$ with an infinity branch $B$. A plane algebraic curve ${\overline{{\cal C}}}$ approaches ${\cal C}$ at $B$ if and only if ${\overline{{\cal C}}}$ has an infinity branch, $\overline{B}$, such that $B$ and $\overline{B}$ are convergent.

**Proof:** Suppose that ${\overline{{\cal C}}}$ approaches ${\cal C}$ at $B$. Then, there exists a leaf $L=\{(z,r(z))\in {\Bbb C}^2:\,z\in {\Bbb C},\,|z|>M\}\subset B$ such that $\lim_{z\rightarrow\infty}d((z,r(z)),\overline{\cal C})=0.$ In addition, let $P=(1:m:0)$ be the infinity point associated to $B$, and let $\{z_n\}_{n\in \mathbb{N}}$ be a sequence in ${\Bbb C}$ such that $\lim_{n\rightarrow\infty} z_n= \infty$. We have that $\lim_{n\rightarrow\infty}d((z_n,r(z_n)),\overline{\cal C})=0$ which implies that $$\lim_{n\rightarrow\infty}d((z_n,r(z_n)),(p_n,q_n))=0,$$ where, 