1. Find all pairs of real numbers ( $\mathrm{x}, \mathrm{y}$ ), for which the equality $\sqrt{x^{2}+y^{2}-1}=1-x-y$ holds.
Answer: All pairs of numbers $(1, t),(t, 1)$, where $\mathrm{t}$ is any non-positive number.

Hint. By squaring and factoring, we get that $\mathrm{x}=1$ or $\mathrm{y}=1$. The square root is non-negative, so the sum $1-x-y \geq 0$. Substituting the possible value of the variable into this inequality, we get that the other variable is non-positive in this case.

Criteria. Not considering that the arithmetic root is non-negative: 3 points.

One set of pairs is missing: 1 point.