1. Determine for which values of $a$ there exist real numbers $x$ and $y$ that satisfy the equation $\sqrt{2 x y+a}=x+y+17$.

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6.I.1. The given equation is equivalent to the system

$$
\left\{\begin{array}{l}
2 x y + a = (x + y + 17)^{2} \\
x + y + 17 \geqslant 0
\end{array}\right.
$$

The first equation can be transformed into the form \((x + 17)^{2} + (x + 17)^{2} = a + 17^{2}\), which, for \(a > -289\), represents the equation of a circle with center at point \(P(-17, -17)\) and radius \(r = \sqrt{a + 289}\). The solutions to the system are the coordinates of the points on this circle that lie in the half-plane defined by the inequality \(x + y + 17 \geqslant 0\). Thus, the system has a solution if and only if the radius of the circle is not less than the distance \(d\) from point \(P\) to the line \(x + y + 17 = 0\) (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-100.jpg?height=662&width=625&top_left_y=771&top_left_x=721)

Since this distance is half the diagonal of a square with side length 17, we have \(d = \frac{17}{\sqrt{2}}\). Therefore, the system has a solution if \(\sqrt{a + 289} \geqslant \frac{17}{\sqrt{2}}\), or \(a + 289 \geqslant \frac{289}{2}\), which implies that \(a \geqslant -\frac{289}{2}\).

Answer: \(a \geqslant -\frac{289}{2}\).