1. Find all real numbers $a$ and $b$

$$
\left\{\begin{array}{l}
\left|\frac{x^{y}-1}{x^{y}+1}\right|=a \\
x^{2}+y^{2}=b
\end{array}\right.
$$

given that $x>0$, there is a unique solution.
Solution. Let $S \subseteq \mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ be the set of pairs $(x, y)$ that satisfy system (1) with $x>0$. From $x^{2}+y^{2}=b$ it follows that $b>0$. If $(x, y) \in S$, then $(x,-y) \in S$, which means that for system (1) to have a unique solution, given $x>0$, it must be $y=0$ and $a=0$. For $a=0$ and $b>1$ the system (1) has solutions $(\sqrt{b}, 0),(1, \sqrt{b-1}),(1,-\sqrt{b-1})$. Therefore, system (1) has a unique solution, given $x>0$, for $a=0$ and $0<b \leq 1$, and the solution is $(\sqrt{b}, 0)$.