(2) (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 that the area $S$ of $\triangle A O B$ is a constant;
(2) Suppose the line $l: y=-2 x+4$ intersects the circle $C$ at two distinct points $M$ and $N$, and $|O M|=|O N|$, find the standard equation of the circle $C$.
(2) (1) Since circle $C$ passes through the origin $O$, we have $|O C|^{2}=t^{2}+\frac{4}{t^{2}}$. Therefore, the equation of circle $C$ is
$$
(x-t)^{2}+\left(y-\frac{2}{t}\right)^{2}=t^{2}+\frac{4}{t^{2}},
$$

which simplifies to $x^{2}+y^{2}-2 t x-\frac{4}{t} y=0$. Thus, $A(2 t, 0)$ and $B\left(0, \frac{4}{t}\right)$. Hence, $S=\frac{1}{2}|O A| \cdot|O B|=\frac{1}{2}|2 t| \cdot\left|\frac{4}{t}\right|=4$, which is a constant.
(2) Since $|O M|=|O N|$ and $|C M|=|C N|$, the line $O C$ is the perpendicular bisector of segment $M N$.
Therefore, $k_{O C}=-\frac{1}{k_{M N}}=\frac{1}{2}$, so the equation of line $O C$ is $y=\frac{1}{2} x$.
Thus, $\frac{2}{t}=\frac{1}{2} t$, solving for $t$ gives $t= \pm 2$.
When $t=2$, the center of the circle $C(2,1)$, and the radius $|O C|=\sqrt{5}$. 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 circle $C$ is $(x-2)^{2}+(y-1)^{2}=5$.