A conjecture involving prime numbers and circles

Assuming Polignac's conjecture we will always be able to find two primes $(c,d)$ such that

$$ \left\lfloor{ \frac{a}{10}} \right\rfloor - \left\lfloor{\frac{b}{10}}\right\rfloor = -\left(\left\lfloor{\frac{c}{10}}\right\rfloor - \left\lfloor{\frac{d}{10}}\right\rfloor\right) $$

(the distance between $a$ and $b$ along the $x$-axis is equal to the negative of the distance between $c$ and $d$) and

$$ a = b,\; \; c=d \; \mod 10 $$

($a$ and $b$, and $c$ and $d$, end in the same digits).

This defines an isosceles trapezium, which is always a cyclic quadrilateral (a quadrilateral such that a circle can be drawn with its 4 vertices.

If $a = b \mod 10 \;$, the above argument still probably holds, but I have not found a proof.

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.