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
$$R^2 = \frac{1}{4}(x_1^2 + x_2^2 + x_3^2 + x_4^2 + \frac{1}{x_1^2} + \frac{1}{x_2^2} + \frac{1}{x_3^2} + \frac{1}{x_4^2}). \quad (1)$$
Proof. Let $(x-a)^2 + (x-b)^2 = R^2$ be the equation of $S$. Then $x_i$ are the roots of the equation $x^4 - 2a x^3 + (a^2 + b^2 - R^2)x^2 - 2b x + 1 = 0$. 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.