$2.20$ ** A square with side length $2 a$ is always located in the first quadrant of the $X Y$ plane. As the square moves, two of its adjacent vertices always remain on the $x$-axis and the $y$-axis, respectively. Find: the locus of the center of the square.
Let $A$ and $B$ be two adjacent vertices of a square, located on the $x$-axis and $y$-axis, respectively. Let $C$ be the center of the square, and draw perpendiculars from $C$ to the $x$-axis and $y$-axis, with $D$ and $E$ being the feet of the perpendiculars. Since $\angle A C B = \angle D C E = 90^{\circ}$, it follows that $\angle A C D = \angle B C E$. Also, $C B = C A$, so Rt $\triangle A C D \cong$ Rt $\triangle B C E$. Therefore, $C D = C E$. This means that the $x$-coordinate and $y$-coordinate of point $C$ are equal. Thus, when the square moves according to the given conditions, its center $C$ moves back and forth along the line segment from $(a, a)$ to $(\sqrt{2} a, \sqrt{2} a)$. Therefore, the trajectory equation of point $C$ is $y = x \quad (a \leqslant x \leqslant \sqrt{2} a)$.