(20 points) In the Cartesian coordinate system, a circle with center $C\left(t, \frac{2}{t}\right)$ passes through the origin $O$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$ (different from the origin $O$).
(1) Prove: The area $S$ of $\triangle A O B$ is a constant;
(2) Suppose the line $l: y=-2 x+4$ intersects $\odot C$ at two distinct points $M$ and $N$, and $|O M|=|O N|$, find the standard equation of $\odot C$.
(1) From the given, the equation of $\odot C$ is
$$
(x-t)^{2}+\left(y-\frac{2}{t}\right)^{2}=t^{2}+\frac{4}{t^{2}} \text {, }
$$

which simplifies to $x^{2}+y^{2}-2 t x-\frac{4}{t} y=0$.
Therefore, $A(2 t, 0), B\left(0, \frac{4}{t}\right)$.
Thus, $S=\frac{1}{2}|O A||O B|$
$=\frac{1}{2}|2 t|\left|\frac{4}{t}\right|=4$ (a constant).
(2) Since $|O M|=|O N|,|C M|=|C N|$, the line $O C$ is the perpendicular bisector of segment $M N$.
Therefore, $k_{o C}=-\frac{1}{k_{M N}}=\frac{1}{2}$.
Thus, $l_{o c}: y=\frac{1}{2} x$.
Hence, $\frac{2}{t}=\frac{1}{2} t \Rightarrow t= \pm 2$.
When $t=2$, the center of the circle $C(2,1)$, and the radius $|O C|=\sqrt{5}$, at this time, the distance from the center $C$ to the line $l$ is $d=\frac{1}{\sqrt{5}}\sqrt{5}$, which does not meet the requirement.
In conclusion, the equation of $\odot C$ is
$$
(x-2)^{2}+(y-1)^{2}=5 \text {. }
$$