Let $d_1\le d_2\le \ldots \le d_{n-1}$ be the distances from $O$ to the center of each disc $D_i$, arranged in increasing order. Then, (b) yields $d_i \le i+1\,,\, i=1,2,\ldots, n-1$. The angle, the disc $D_i$ is seen from $O$ is at least $2/d_i$. Assign to each disc $D_i$ the set: \[A_i=\{\varphi \in [0.2\pi) \mid \text{ the line through } O \text{ and argument } \varphi\ \text{ stabs } D_i \}\] Thus, we have $A_i \subset [0,2\pi)\,,\, m(A_i) \ge 4/d_i$, where $m(A_i)$ is the length of $A_i$ (or rather the sum of lengths of its components ). There are at least $\left(\sum m(A_i)\right)/2\pi$ sets among $A_i$ with a common point. We have: \[\sum m(A_i) \ge 4\sum_{i=2}^n \frac{1}{i}\ge 4\left(\ln (n+1) -1\right) \] It means there exist at least $\frac{2}{\pi}\left( \ln(n+1)-1\right)$ sets among $A_i$ with a common point $\varphi$. Then the line $\ell$ through $O$ with $\angle(O\ell, Ox)=\varphi $ stabs all corresponding discs. EDIT: The condition (a) is redundant.