Write a non-trivial element of $ H^2 (\mathbb{C}P^2 \backslash \{ x = 0 \})$ I am trying to understand a bit of the homology of a projective space minus a hyperplane.  
$$ \mathbb{C}P^2 \backslash \{ x = 0\} = \big\{ [x:y:z]: x \neq 0\big\} / \mathbb{C}^\times $$
This is topological space and therefore we can ask about finding non-trivial 2-cycles in this space.  
One can ask more generally about the topology of projective spaces with hyperplanes removed.  If you remove a hyperplane from $\mathbb{R}^3$ the space $\mathbb{R}^3 \backslash \{ x = 0\}$ is no longer connected.  However, there is a non-trivial 1-cycle in Euclidean 4-space with a single hyperplane removed  $S_1 \subseteq \mathbb{R}^4 \backslash \{ x =y= 0\}$ :
$$ \big[ \big\{ ( \cos \theta,\sin \theta,0,0) : \theta \in [0,2\pi]  \big\} \big]  \in H_1 \big( \mathbb{R}^4 \backslash \{ x=y=0 \} \big) $$
Complex projective space is also a 4-manifold, since it's the quotient of $\big(\mathbb{C}^3 \backslash \{0\} \big) / \mathbb{C}^\times$ the dimension is $6 - 2 = 4$ (or complex dimension $3 - 1 = 2$.  I guess the confusing part is that:
$$  \mathbb{C}^3 \backslash \{ x = 0\} \text{ is connected} $$
Back to my original space, I am looking for non-trivial loops in $\mathbb{C}P^2 \backslash \{ x = 0\}$ that are not already in $\mathbb{C}P^2$.  We have for projective space:
$$ [hyperplane] \in H_2(\mathbb{C}P^2, \mathbb{Z}) \simeq \mathbb{Z} $$
My question is how the topology changes upon removing this hyperplane.  I  am flipping through Hatcher's Topology textbook, which says in Chapter 2 that Homology behaves nicely under excision.   However, the nature of my question is elementary.
Perhaps I can phrase it this way... what are the generators of the homology of projective space minus a hyperplane:
$$ H^2 \big( \mathbb{C}P^2 \backslash \{ x = 0\}\big)$$
I'm not just looking for the homology group, but the topological subspaces that generate them.

The map
$$
\begin{align*}
\mathbb{C}P^2\backslash\{x=0\}&\to\mathbb{C}^2,\\
[x:y:z]&\mapsto \left(\frac{y}{x},\frac{z}{x}\right)
\end{align*}
$$
is a homeomorphism. This means the (co)homology groups of $\mathbb{C}P^2\backslash\{x=0\}$ are the same as those of $\mathbb{C}^2$, so in particular $H^2(\mathbb{C}P^2\backslash\{x=0\})$ is trivial.
