Given the rhombus $ABCD\ :\ A(a,0),B(0,b),C(-c,0),D(0,-b)$. The circle $(x-a)^{2}+y^{2}=a^{2}+b^{2}$ intersects the line $CD\ :\ y=-\frac{b}{c}(x+c)$ in the point $E(\frac{2c(ac-b^{2})}{b^{2}+c^{2}},-\frac{b(2ac-b^{2}+c^{2})}{b^{2}+c^{2}})$. The line $BE$ intersects the x-axis in the point $S(\frac{ac-b^{2}}{a+c},0)$. The circle through $A,S,D,E\ :\ x^{2}+y^{2}-\frac{a^{2}+2ac-b^{2}}{a+c} \cdot x+\frac{c(a^{2}+b^{2})}{b(a+c)} \cdot y +\frac{a(ac-b^{2})}{a+c}=0$.

Complex Bash EzPz Set $a = 0$, $b = 1$. Let $d$ be on the unit circle alongside $b$ and $e$. We have $c = d + 1$. checking the intersection of cd with the unit circle we find $e = \frac{-1}{d}$. By checking for the intersection we find $s = \frac{d}{d + 1}$ We now check that $a, s, d, e$ are cyclic by checking that: \[ T := \frac{(s - a) \div (e - a)}{(s - d) \div (e - d)} = d + \frac{1}{d} \] is a real number, which is clear.