Given the values of $a$ and $b$. Solve the following system of equations:

$$
\frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=a, \quad \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=b
$$
By squaring the equations and then taking their differences, we arrive at the following relationship:

$$
a^{2}-b^{2}=x^{2}-y^{2}
$$

Substituting this into (1), we obtain the following system of equations:

$$
\begin{aligned}
& x-y \sqrt{a^{2}-b^{2}}=a \sqrt{1-a^{2}+b^{2}} \\
& y-x \sqrt{a^{2}-b^{2}}=b \sqrt{1-a^{2}+b^{2}}
\end{aligned}
$$

Solving this, we get:

$$
x=\frac{a+b \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}}, \quad y=\frac{b+a \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}}
$$

These values exist only if the quantities under the square roots are non-negative and the denominator is not zero, i.e., if $a$ and $b$ satisfy the condition $0 \leq a^{2}-b^{2}<1$. In this case, (2) is also satisfied, so the $x$ and $y$ under (3) give the unique solution to the system of equations. If $a^{2}-b^{2}<0$ or $a^{2}-b^{2} \geq 1$, then (1) has no solution.

(K. M.)

Magyar Ákos (Budapest, Fazekas M. Gyak. Gimn., I. o. t.)