Belarusian Mathematical Olympiad

Let the abscissae of the points $A$, $B$, $C$, $D$, $F$ and $G$ be $a$, $b$, $c$, $d$, $f$ and $g$ respectively. If $O_1(\alpha ,\beta )$ is the center of the circle $S_1$, the coordinates of the points $A$, $B$, $C$ and $D$ satisfy the system of equations $$\{\begin{array}{l}(x-\alpha {)}^2+(y-\beta {)}^2=R^2,\\ y=\frac{1}{x}.\end{array}$$ Eliminating $y$ we obtain the equation $$x^4-2\alpha x^3+({\alpha }^2+{\beta }^2-R^2)x^2-2\beta x+1=0,$$ with solutions $a$, $b$, $c$ and $d$. Whence $a+b+c+d=2\alpha$ and $abcd=1$. Similarly, if $O_2(\lambda ,\mu )$ is the center of the circle $S_2$, then $a+b+f+g=2\lambda$ and $abfg=1$. Since the radii of the circles $S_1$ and $S_2$ are equal, the quadrilateral $A{O}_{1}B{O}_2$ is a rhombus, hence the midpoints of $AB$ and $O_1{O}_2$ coincide, therefore $\frac{a+b}{2}=\frac{\alpha +\lambda }{2}$ which gives $a+b=\alpha +\lambda$. So $$2\alpha +2\lambda =(a+b+c+d)+(a+b+f+g)=2\alpha +2\lambda +c+d+f+g,$$ whence $f+g=-(c+d)$. Moreover, $cd=fg=(ab{)}^{-1}$. Rewrite it as $f+g=(-c)+(-d)$ and $f\cdot g=(-c)\cdot (-d)$. Thus the pairs $(f,g)$ and $(-c,-d)$ have equal sums and products. So either $f=-c$ and $g=-d$ or $f=-d$ and $g=-c$. If, for example, $f=-c$ and $g=-d$, then the pairs $F,C$ and $G,D$ of points are symmetric with respect to the point of origin, i.e. $FGCD$ is a parallelogram with the center at the origin.