Selection and Training Session

Answer: $\sqrt{R_1^2+R_3^2-R_2^2}$.

First we prove the following

Lemma. Let $x_1,x_2,x_3,x_4$ be abscissae of the intersection points of a circle with the hyperbola $y=\frac{1}{x}$. Then $$R^2=\frac{1}{4}(x_1^2+x_2^2+x_3^2+x_4^2+\frac{1}{x_1^2}+\frac{1}{x_2^2}+\frac{1}{x_3^2}+\frac{1}{x_4^2}).(1)$$ Proof. Let $(x-a{)}^2+(x-b{)}^2=R^2$ be the equation of $S$. Then $x_i$ are the roots of the equation $x^4-2a{x}^3+(a^2+b^2-R^2)x^2-2bx+1=0$. From Viète's formulas it follows that $\sum x_i=2a$, $\prod x_i=1$, $\sum x_i{x}_j=a^2+b^2-R^2=\sum \frac{1}{x_i{x}_j}$, $\sum x_i{x}_j{x}_k=\sum \frac{1}{x_k}=2b$. From this system of equalities we can easily express $R^2$ and obtain (1).

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Now let $Z_1,Z_2,Z_3,Z_4$ and $Z_5,Z_6,Z_7,Z_8$ be the abscissae of the intersection points of the hyperbola with $S_1$ and $S_3$ respectively. Then $R_1^2+R_3^2=\frac{1}{4}({\sum }_{i=1}^8{Z}_i^2+{\sum }_{i=1}^8\frac{1}{Z_i^2})=R_2^2+R_4^2$ whence follows the answer.