9. Given $\triangle A B C, \angle C=90^{\circ}, A, B$ points slide on the positive half of the $x$-axis and the positive half of the $y$-axis, respectively. Find the equation of the locus of point $C$.
9. Solution: Let $\angle ABC = \alpha, AC = b, BC = a$. Without loss of generality, assume $b \geqslant a$ (i.e., $\alpha \geqslant 45^{\circ}$). Connect $OC$. Since $\angle BCA + \angle AOB = 90^{\circ} + 90^{\circ} = 180^{\circ}$, points $O, A, C, B$ are concyclic, and $\angle AOC = \angle ABC = \alpha$. Point $C$ lies on the line $y = \frac{b}{a} x$ (where $\tan \alpha = \frac{b}{a}$). However, not all points on this line meet the requirements; the desired locus is a segment $C_1C_2$. Here, $C_1$ is at a distance $a$ from the origin, so its coordinates are $\left(\frac{ab}{c}, \frac{b^2}{c}\right)$, where $c$ is the hypotenuse length. The coordinates of $C_2$ are $(a, b)$. Therefore, the equation of the locus should be written as $y = \frac{b}{a} x, \left(\frac{ab}{c} \leqslant x \leqslant a\right)$.