TITLE: A conjecture involving prime numbers and circles
QUESTION [51 upvotes]: Given the series of prime numbers greater than $9$, we organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.

My conjecture is:

Given any two primes (i.e. given any two points in the above scheme), it is always possible to find a circle passing through at least other two points, representing other two primes.

Here I present some examples, taking two random points. Sorry for the bad quality of the picture.

Since I am not an expert of prime numbers, this can be an obvious result (if true, of course). In this case, I apologize for the trivial question. 
However, I tried to attack the problem by means of the equation of the circle, but I got lost. Thanks for your help!
NOTE: You might be interested in this and in this other post. Also, here I state a similar conjecture for ellipses.

REPLY [5 votes]: Here is some loose intuition to convince you that it is equally hard as the twin prime conjecture. Especially, to convince you that there's no point in trying to prove or disprove it:

At most as hard as the twin prime conjecture:
Take two primes $p_1,p_2$. If the twin prime conjecture is true, it is reasonable to expect that, for any even $2k \geq 2$ and $n \bmod 10$ there are infinitely many prime pairs $(q_1,q_2)$ with $q_2-q_1 = 2k$ and $q_1 \equiv n \pmod{10}$.1 Then for any given $p_1,p_2$ not congruent mod $10$ we can find two other primes to form a trapezium. This takes care of the case where $p_1,p_2$ are not congruent, at least.
At least as hard as the twin prime conjecture:
Four points with coordinates $(x_i,y_i)$ are concyclic iff 
$$\begin{vmatrix}
1 & x_1 & y_1 & x_1^2 + y_1^2 \\
1 & x_2 & y_2 & x_2^2 + y_2^2 \\
1 & x_3 & y_3 & x_3^2 + y_3^2 \\
1 & x_4 & y_4 & x_4^2 + y_4^2
\end{vmatrix} = 0$$
This gives, for every pair of primes $(p_1,p_2)$ a degree $4$ equation in two primes $q_1,q_2$ (and their residues mod $10$). Current methods are nowhere near proving that it has a solution; indeed, we cannot even show that the degree 1 (!) equation $$q_2-q_1-2k = 0$$ has a solution for every $k$.


1 Although, there was an article that appeared a few years ago with some computations, suggesting that the distribution of the remainder of three consecutive primes mod a given integer, is not uniform. Anyway.