Set of balls which the number of the ball intersects lines on the plane is bounded Does there exist the set of balls(may be not disjoint) $X=\{B_i\subset\mathbb{R^2};i\in I\}$, satisfing following properties?(Note that the ball has a positive real radius) 


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*Let the set of all lines in plane to be $L$.  For each  $l \in L,\ l\cap B_i \ne \varnothing $ for some $i\in I$.

*$\{N_l=\text{the number of balls which the line} \ l \ \ \text{meets}\mid l\in L\}$  is bounded. 


(This does not mean the number of ball which line meets is finite. This statement is stronger.)  
Is there such a set??
Edit: I exchage "transversal" with "meet".
Here is a original problem posted in Math Stack Exchange. https://math.stackexchange.com/questions/1387435/there-is-a-no-set-which-every-line-meets-the-ball
 A: No, such $X$ does not exist.
Assume the contrary. Take a point $O$ which is outside all disks and a circle $\omega$ centered at $O$. We implement polarity with respect to this circle (thus, in what follows we consider only lines not passing through $O$).
Take any ball $B_i\in X$. The poles of all lines tangent to $B_i$ form a hyperbola $h_i$ with a focus at  $O$; then the poles of all lines intersecting $B_i$ form the set $H_i$ consisting of $h_i$ and of the interior points of its two branches. Let $A_i$ be the cone bounded by the asymptotes of $h_i$ and containing $H_i$ (we assume that $A_i$ contains no points of the asymptotes except for $O\in A_i$). Notice here that in any closed subcone of $A_i$, all points which are sufficiently far from $O$ lie in $H_i$.
The fact that $X$ satisfies the required properties yields that the sets $H_i$ cover the plane (except for $O$) but in a bounded number of layers. We will show that this is impossible.
Take an arbitrary $A_{i_1}$ and some its closed subcone $S_1$. Some interior point $x_1$ of $S_1$ is not covered by $H_{i_1}$, so it is covered by some $H_{i_2}$. Take a closed subcone $S_2$ of $S_{i_1}\cap A_{i_2}$ and proceed similarly. Thus, on the $(k+1)$th step we choose an interior point $x_k\in S_k$ not covered by $H_{i_1},\dots,H_{i_k}$ (all points close enough to $O$ are such), take $H_{i_{k+1}}$ covering $x_k$, and take a closed subcone $S_{k+1}$ of $S_k\cap A_{i_{k+1}}$.
Finally, for every $k$, all points in $S_{k}$ which are sufficiently far from $O$ are covered by each of $H_{i_1},\dots,H_{i_k}$, so some point is covered by arbitrarily many of the $H_i$.
