Example 6 Suppose a circle satisfies: (1) the length of the chord intercepted on the $y$-axis is 2; (2) it is divided into two arcs by the $x$-axis, with the ratio of their arc lengths being $3: 1$. Among all circles satisfying (1) and (2), find the equation of the circle whose center is closest to the line $l: x-2y=0$.
Let the center of the required circle be $P(a, b)$, and the radius be $r$, then the distances from $P$ to the $x$ and $y$ axes are $|b|,|a|$ respectively.
From the fact that the central angle of the circle intercepted by the $x$-axis is $90^{\circ}$, the length of the chord intercepted by the $x$-axis is $\sqrt{2} r$, hence $r^{2}=2 b^{2}$.
Also, the length of the chord intercepted by the $y$-axis is 2, so we have $r^{2}=a^{2}+1$, thus $2 b^{2}-a^{2}=1$.
Let the distance from $P(a, b)$ to the line $x-2 y=0$ be $d$, then $d=\frac{|a-2 b|}{\sqrt{5}}, 5 d^{2}=(a-2 b)^{2}=a^{2}-4 a b+4 b^{2}$ $\geqslant a^{2}+4 b^{2}-2\left(a^{2}+b^{2}\right)=2 b^{2}-a^{2}=1$.
Equality holds if and only if $a=b$, at this time $\left\{\begin{array}{l}a=1 \\ b=1\end{array}\right.$ or $\left\{\begin{array}{l}a=-1, \\ b=-1 .\end{array}\right.$
Therefore, the equations of the required circles are $(x-1)^{2}+(y-1)^{2}=2$ or $(x+1)^{2}+(y+1)^{2}=2$.