Example 1 Given a curve $C: x^{2}+y^{2}-2 x-2 y+1=0$ and a tangent line $l$ that intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. $O$ is the origin, $|O A|=a,|O B|=b, a>2$, $b>2$.
(1) Prove: $(a-2)(b-2)=2$;
(2) Find the equation of the locus of the midpoint of segment $A B$;
(3) Find the minimum value of the area of $\triangle A O B$.
Solution: (1) From the problem, the equation of line $l$ is $b x+a y-a b=0$.
Curve $C$ can be simplified to $(x-1)^{2}+(y-1)^{2}=1$, which represents a circle with center $(1,1)$ and radius 1.
The necessary and sufficient condition for line $l$ to be tangent to the circle is
$$
1=\frac{|a+b-a b|}{\sqrt{a^{2}+b^{2}}} \text {, i.e., }(a-2)(b-2)=2 \text {. }
$$
(2) Let the midpoint of $A B$ be $M(x, y)$, then by the midpoint coordinate formula we get $a=2 x, b=2 y$, so according to conclusion (1) we have $(x-1)(y-1)=\frac{1}{2}$ (where $x>1, y>1$).
$$
\text { (3) } \begin{aligned}
S_{\triangle A O B} & =\frac{1}{2} a b=\frac{1}{2}[-2+2(a+b)] \\
& =(a-2)+(b-2)+3 \\
& \geqslant 2 \sqrt{(a-2)(b-2)}+3 \\
& =2 \sqrt{2}+3 .
\end{aligned}
$$

Therefore, the minimum value of the area of $\triangle A O B$ is $2 \sqrt{2}+3$.