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) 


1. Let the set of all lines in plane to be $L$. For each $l \in L,\ l\cap B\_i \ne \varnothing $ for some $i\in I$.
2. $\{N\_l=\text{the number of balls which the line} \ l \ \ \text{meets}\mid l\in L\}$ 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>



Set of balls which the number of the ball intersects lines on the plane is bounded
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 $H\_i$.


