Circles on the plane such that every line intersects at least one of them but no line intersects more that 100 of them I have a serious problem with this problem: Is it possible to  Draw circles on the plane such that every line intersects at least one of them but no line intersects more that 100 of them !?
Any help or suggestion would be helpful. 
 A: I have only a partial (negative) solution. 
I can prove that there is no chain of circles of infinite length. That is, no sequence of circles without repetitions, where every two consecutive circles intersect or touch and the sum of the radii of the circles diverges. In particular, this rules out the suggestions with hyperbolas or parabolas, if I understand them correctly.
Proof by contradiction. Let $C$ be one of the circles of the chain and let $P$ be its center. We will prove that some ray starting in $P$ intersects infinitely many circles. Let's call $P$-rays the rays starting at $P$.
A circle of radius $r$ with center at distance $\ell$ from $P$ intersects at least $2r/(\ell\cdot 2\pi)$ fraction of the $P$-rays. (This is not true if $r>\ell\cdot \pi$, of which we take care at the end.)
Let $r_i$ be the sequence of radii of the circles along the chain starting with $C$. Then the distance $\ell_i$ between $P$ and the center of the $i$-th circle is at most $2(r_1+r_2+\cdots + r_i)$. So the $i$-th circle blocks at least $(1/(2\pi)) \cdot r_i/(r_1+\cdots +r_i)$ fraction of $P$-rays. Since the chain has infinite length, the sum $r_1 + r_2 + \cdots$ diverges. Then also the sum of $r_i$ divided by partial sums diverges. That is, the sum of fractions of blocked $P$-rays goes to infinity and some $P$-ray is blocked infinitely many times.
We still need to do something with the circles of the chain with $r>\ell\cdot \pi$. First of all, $C$ is one of them: we counted that it blocks $(1/(2\pi)) \cdot r_1/r_1$ fraction, and that's true. For others, we may have counted that they block many times more than 100% percent of the $P$-rays. But since each such circle contains $P$ and crosses every $P$-ray, there are at most $99$ of such circles. So the excess counted to the sum is finite and the corrected sum still diverges.
A: This is impossible.
$\bf{Proof}$: by contradiction. Suppose it were possible to draw circles on the plane such that every line intersects at least one of them but no line intersects more that 100 of them. Denote $\mathbb{R}/\pi\mathbb{R}$ with $\mathbb{R}_\pi$, choose a point $(x_0,y_0)$ on the plane, and define $c_{(x_0,y_0)}:\mathbb{R}_\pi\to\mathbb{Z}_{\leq100}$ such that $c_{(x_0,y_0)}(\theta)$ be the number of circles which the line $(y-y_0)=\tan(\theta)(x-x_0)$ intersects (or the line $x=x_0$ if $\theta=\frac{\pi}{2}$). $c_{(x_0,y_0)}$ must reach a maximum for every open interval $(a,b)\subseteq\mathbb{R}_\pi$ since it is positive and integer valued, say this maximum is reached at $\phi\in (a,b)$. Let $S_\phi$ denote the set of circles which the line $(y-y_0)=\tan(\phi)(x-x_0)$ intersects, and $f(C)$ be the set of angles $\theta$ such that $(y-y_0)=\tan(\theta)(x-x_0)$ intersects circle $C$. Define $M(a,b)$ as the open set $$(a,b)\cap\bigcap_{C\in S_\phi}f(C)$$ with endpoint removed if closed.
Since $\phi$ maximizes $c_{(x_0,y_0)}$ on $(a,b)$, and every line going through $(x_0, y_0)$ with angle in $M(a,b)$ must go through all the circles which the line with angle $\phi$ goes through (and cannot go through any other circles as $\phi$ is the maximum), the open cone of lines $(y-y_0)=\tan(\theta)(x-x_0)$, $\theta\in M(a,b)$ must contain a finite number of circles. Let $$B_{(x_0,y_0)}=\bigcup_{a,b\in \mathbb{R}/\pi\mathbb{R}}M(a,b)$$ I originally attempted to demonstrate that $B_{(x_0,y_0)}=\mathbb{R}_\pi$ and then, since $\mathbb{R}_\pi$ is compact, find a finite open cover and deduce a finite number of circles, but failed ($B_{(x_0,y_0)}=\mathbb{R}_\pi$ isn't always a true statement). This obstacle can however be surmounted.
One can demonstrate that $|\mathbb{R}_\pi-B_{(x_0,y_0)}|$ cannot be infinite (again, by contradiction). Suppose it were. Since the set is an infinite subset of a compact set, it must have a limit point in $\mathbb{R}_\pi$, which we denote with $\theta_l$, and let $\theta_1<\theta_2<...<\theta_n<...$ all be in $\mathbb{R}_\pi-B_{(x_0,y_0)}$ and converge to $\theta_l$. Pick some $C_1\in S_{\theta_l}$. There must be some $\theta_{m_1}\in f(C_1)$ since $\theta_i$ converges to $\theta_l$. However, since $\theta_{m_1}\not\in B_{(x_0,y_0)}$, there must exist another circle $C_2\in S_{\theta_l}$ such that $\theta_{m_1}\not\in f(C_2)$, and there must be some $\theta_{m_2}\in f(C_2)$. We repeat this process to find a $C_3$ distinct from $C_1$ and $C_2$, and so on. Since this can be iterated infinitely many times as there are infinitely many terms in the sequence which converges to $\theta_l$, we conclude that we can find infinitely many distinct circles in $S_{\theta_l}$, which is a contradiction. Therefore, there can be only a finite number of elements in $\mathbb{R}_\pi-B_{(x_0,y_0)}$.
Let $L_{(x,y)}(A)$ denote the set of lines going through the point $(x,y)$ at an angle $\theta\in A$, and pick $(x_1,y_1)\not\in L_{(x_0,y_0)}(\mathbb{R}_\pi-B_{(x_0,y_0)})$. There exists compact subsets $\overline{B}_{(x_0,y_0)}\subset B_{(x_0,y_0)}$ and $\overline{B}_{(x_1,y_1)}\subset B_{(x_1,y_1)}$ such that $$L_{(x_0,y_0)}(\mathbb{R}_\pi-\overline{B}_{(x_0,y_0)})\cap L_{(x_1,y_1)}(\mathbb{R}_\pi-\overline{B}_{(x_1,y_1)})$$ is bounded since $\mathbb{R}_\pi-B_{(x_0,y_0)}$ and $\mathbb{R}_\pi-B_{(x_1,y_1)}$ are finite. Furthermore, since $\overline{B}_{(x_0,y_0)}$ and $\overline{B}_{(x_1,y_1)}$ admit finite covers with respective sets of the form $M(a,b)$ as described above, the number of circles intersecting with  $$L_{(x_0,y_0)}(\overline{B}_{(x_0,y_0)})\cup L_{(x_1,y_1)}(\overline{B}_{(x_1,y_1)})$$ is finite, and the circles are thus bounded. As a result, the set of circles in $$\mathbb{R}^2=\Big{(}L_{(x_0,y_0)}(\mathbb{R}_\pi-\overline{B}_{(x_0,y_0)})\cap L_{(x_1,y_1)}(\mathbb{R}_\pi-\overline{B}_{(x_1,y_1)})\Big{)}\cup\Big{(}L_{(x_0,y_0)}(\overline{B}_{(x_0,y_0)})\cup L_{(x_1,y_1)}(\overline{B}_{(x_1,y_1)})\Big{)}$$ is bounded, which yields a contradiction as there are lines which lie outside the bounded area (and therefore do not intersect any circle).
Note: I do not claim credit for this proof; I derived part of it then found that a more formal and complete version of my ideas had already been explored here. Any detail which I may have overlooked is likely addressed on that page.
