Consider all pairs of numbers $(x, y)$ that satisfy the equation

$$
x^{2} y^{2}+x^{2}-10 x y-8 x+16=0
$$

What values can the product $x y$ take?
The equation can be written in the form

$$
(x y-5)^{2}+(x-4)^{2}=25
$$

By substituting $x y=z$, we arrive at the equation of a circle, where $z$ takes values in the interval $[0 ; 10]$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6dea220aae48ea59e2c5g-1.jpg?height=536&width=496&top_left_y=377&top_left_x=804)

If $x \neq 0$, then every real number, and thus $z$, can be expressed in the form $x y$, so all $z$ for which there exists an $x \neq 0$ such that

$$
(z-5)^{2}+(x-4)^{2}=25
$$

belong to the range of $x y$. However, for every

$$
z \in[0 ; 10]
$$

there exists such an $x$, since at $x=0$ and $x=8$, the value of $(x-4)^{2}$ is the same. Therefore, the range of $x y$ is the closed interval $[0; 10]$. Miklós Horváth (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)