Cremona transformation and line arrangements We work over the complex numbers. Let $A_3 \subseteq \mathbb{P}^2 $ be the following arrangement: take three generic lines in the plane and pass an ellipse through
the three intersection points. We know that the following Cremona transformation
$$
f:[x:y:z] \mapsto [yz:xz:xy],
$$
send $A_3$ to $f(A_3)$, when $f(A_3)$ is an arrangement of four generic lines.
Now, let $A_4$ be the following arrangement: take a quadrilateral bounded in a circle, where there is no pair of parallel edges.
Extend the edges till infinity, and complexity this arrangement, to get an arrangement $A_4$ of $4$ lines and a circle. I would like
to find a Cremona transformation (or a composition of the basic Cremona transformations) that will send $A_4$ to an arrangment composed only
of lines. is this possible? if so, what is the resulting arrangement of lines? And what about $A_5$, i.e. starting from a pentagon bounded in a circle? Thank you, Jean. 
 A: First of all, I think that your initial description of what happens with $A_3$ is not completely accurate. If one really takes $A_3$ to be three generic lines and an ellipse through the intersection points, then $f(A_3)$ would not be an arrangement of four lines as you descibe. Indeed, if $C$ is an irreducible curve in $\mathbb{P}^2$ of degree $d$, then the degree of the curve $f(C)$ is given by the formula $2d-m_1-m_2-m_3$, where the $m_i$ are the multiplicities of $C$ at the coordinate points $[1,0,0]$, $[0,1,0]$, and $[0,0,1]$. So a generic line gets mapped to a curve of degree 2 by $f$. 
Of course, we can take any three lines in $\mathbf{P}^2$ to any other three by projective transformations, so let's do that, and assume our lines $L_1$, $L_2$, $L_3$ really are the coordinate lines $x=0$, $y=0$, and $z=0$. (This is probably what you had in mind, but I thought it might be good to clarify for readers of the question.) Then if $C$ is an ellipse passing through the three intersection points, its image $f(C)$ is indeed a line. However, note that the $f(L_i)$ are not lines in $\mathbf{P}^2$: they are points!  For example, $f(L_1)=[1,0,0]$. So we have to be careful about what we mean by the notation $f(A_3)$, since $f$ has actually contracted three of the four curves in the arrangement $A_3$.
Still, your description of the situation is accurate, if interpreted correctly, because $f$ has also "extracted" three lines. More precisely, $f: \mathbf{P}^2 \dashrightarrow \mathbf{P}^2$ is a rational map with the property that the three coordinate lines $\Lambda_1$, $\Lambda_2$, $\Lambda_3$ in the target $\mathbf{P}^2$ do not come from any curves in the domain. And the $\Lambda_i$ together with $f(C)$ do form an arrangement of four lines, and they are generic in the sense that no three intersect at a point. 
What now of $A_4$? Well, one can check that to fulfil your requirements, two of the lines in the arrangement, say $L_1$ and $L_2$, must be be coordinate lines, and the other two, $L_3$ and $L_4$, are lines that pass through just one of the coordinate points. Finally, the circle $C$ passes through each point where two of the $L_i$ intersect. Now three of the four points $L_i \cap L_j$ are coordinate points, so by the degree formula above, $f(C)$ will again be a line. Moreover, since $L_3$ and $L_4$ pass through exactly one coordinate points, their images $f(L_3)$ and $f(L_4)$ are also lines by the degree formula. Finally $L_1$ and $L_2$ are contracted by $f$ as before, but again we get 3 new lines $\Lambda_1$, $\Lambda_2$, $\Lambda_3$. So if we understand $f(A_4)$ to mean the images of the non-contracted lines in $A_4$ together with the "new" lines $\Lambda_i$, then again $f(A_4)$ is an arrangement of lines, but now with one more member than $A_4$ had!
This is long enough already, so let me leave it to you to work out what happens with $A_n$ for higher $n$. I hope this helps.
