13.3 .8 * Given that $\odot C$ satisfies the following three conditions:
(1) $\odot C$ is tangent to the $x$-axis; (2) the center $C$ lies on the line $3x - y = 0$; (3) $\odot C$ intersects the line $x - y = 0$ at points $A$ and $B$, and the area of $\triangle ABC$ is $\sqrt{14}$. Find the equation of $\odot C$ that meets these conditions.
Given that the center $C$ is on the line $3x - y = 0$, let the coordinates of $C$ be $(a, 3a)$. Let the radius of $\odot C$ be $r$. Since $\odot C$ is tangent to the $x$-axis, we have $|3a| = r$. The distance $d$ from $C$ to the line $x - y = 0$ is $d = \frac{|a - 3a|}{\sqrt{2}} = \sqrt{2}|a|$. Since $\odot C$ intersects the line $x - y = 0$ at points $A$ and $B$, we have
$$
\sqrt{2}|a| < r \text{ and } |AB| = 2 \sqrt{r^2 - d^2} = 2 \sqrt{9a^2 - 2a^2} = 2 \sqrt{7}|a|,
$$

Therefore, the area of $\triangle ABC$ is $\frac{1}{2} \times |AB| \times d = \frac{1}{2} \times 2 \sqrt{7}|a| \times \sqrt{2}|a| = \sqrt{14} a^2$. Since the area of $\triangle ABC$ is $\sqrt{14}$, we have $a^2 = 1$, $a = +1$, and $r = 3$. Thus, the equation of the circle $\odot C$ is $(x - 1)^2 + (y - 3)^2 = 9$ or $(x + 1)^2 + (y + 3)^2 = 9$.