Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles. 
Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles.


I have solved this one but now I can't remember how I did it. I just remember that I used some linear algebra and double counting.
Suppose each point $P_i\in \{P_1,P_2,...P_{22}\}$ is on $p_i\geq 7$ circles among circles $C_1,C_2,...,C_{22}$. Since each pair $\{C_i,C_j\}$ share at most $2$ points and each point is on $\displaystyle{p_i\choose 2}$ pair of circles, we have:
$$ 2\cdot {22\choose 2} \geq \sum _{i=1}^{22} {p_i\choose 2} \geq 22 {7\choose 2} $$
Since we must have all equalities we deduce that $p_i = 7$ for all $i$ and each pair of circles intersect in exactly two points. Now since $$ 22\cdot 7 \leq \sum _{i=1}^{22} |C_i| =  \sum _{i=1}^{22}p_i = 22\cdot 7$$
so each circle contains exactly $7$ points.
 A: As we already see in the introduction each point is on $7$ circles and each circle contains $7$ points. Also every two circles meet at $2$ points. 
Let $A$ be an incident matrix, so $a_{ij} = 1$ if $P_j \in C_i$, else $a_{ij} =0$. Let $M:= A\cdot A^T$, then determinant of
$M$ is a perfect square: $\det (M) =\det (A\cdot A^T)
=\det A \cdot \det A^T =(\det A)^2$. But $$ M=
\begin{bmatrix}
  7 & 2 & 2 & \cdots & 2 & 2 \\
  2 & 7 & 2 & \cdots & 2 & 2 \\
  2 & 2 & 7 & \cdots & 2 & 2 \\
  \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
  2 & 2 & 2 & \cdots& 7 & 2 \\
  2 & 2 & 2 & \cdots & 2 & 7 \\
\end{bmatrix} \sim
\begin{bmatrix}
  5 & 0 & 0 & \cdots & 0 & -5 \\
  0 & 5 & 0 & \cdots & 0 & -5 \\
  0 & 0 & 5 & \cdots & 0 & -5 \\
  \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
  0 & 0 & 0 & \cdots& 5 & -5 \\
  2 & 2 & 2 & \cdots & 2 & 7 \\
\end{bmatrix} \sim
\begin{bmatrix}
  5 & 0 & 0 & \cdots & 0 & 0 \\
  0 & 5 & 0 & \cdots & 0 & 0 \\
  0 & 0 & 5 & \cdots & 0 & 0 \\
  \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
  0 & 0 & 0 & \cdots& 5 & 0 \\
  2 & 2 & 2 & \cdots & 2 & 49 \\
\end{bmatrix}
$$
so $\det (M)= 5^{21}\cdot 49$ which is not a perfect square. Thus such a configuration does not exist.
