In the plane rectangular coordinate system $x O y$, the equation of $\odot A$ is $(x-2)^{2}+(y-2)^{2}=1$. Two circles $\odot O_{1}$ and $\odot O_{2}$, both with radius $r$ and externally tangent to each other, are also externally tangent to $\odot A$. Furthermore, $\odot O_{1}$ and $\odot O_{2}$ are tangent to the $x$-axis and $y$-axis, respectively. Find $r$.
Solution: As shown in Figure 2 and Figure 3, since the center of circle $A$, $A(2,2)$, lies on the line $y=x$, and the radii of $\odot O_{1}$ and $\odot O_{2}$ are equal, the entire figure is symmetric about the line $y=x$. Therefore, the tangent point $T$ of $\odot O_{1}$ and $\odot O_{2}$ lies on the line $y=x$. Let $\odot O_{1}$ and $\odot O_{2}$ be tangent to the $x$-axis and $y$-axis at points $Q$ and $P$, respectively. The lines $Q O_{1}$ and $P O_{2}$ intersect at point $H$, which lies on the line $y=x$. The quadrilateral $O Q H P$ is a square. Connect $O_{1} A$.
It is easy to see that $O_{1} A=1+r, O_{1} T \perp T A$,
$$
T A=\sqrt{(1+r)^{2}-r^{2}}=\sqrt{1+2 r} .
$$

Therefore, $O Q=O T=O A \mp T A=2 \sqrt{2} \mp \sqrt{1+2 r}$.
(Figure 2 takes the negative sign, Figure 3 takes the positive sign).
On the other hand,
$$
T H=T O_{1}=r, O_{1} H=\sqrt{2} r, H Q=(1+\sqrt{2}) r .
$$

By $H Q=O Q$ we get
$$
(1+\sqrt{2}) \cdot r=2 \sqrt{2} \mp \sqrt{1+2 r} \text {, }
$$

which is
$$
\pm \sqrt{1+2 r}=2 \sqrt{2}-(1+\sqrt{2}) r \text {. }
$$

Squaring both sides of the above equation, we get
$$
(3+2 \sqrt{2}) r^{2}-(10+4 \sqrt{2}) r+7=0 \text {. }
$$

Solving for $r$ yields $r=\frac{5+2 \sqrt{2} \pm \sqrt{12+6 \sqrt{2}}}{3+2 \sqrt{2}}$
$$
=7-4 \sqrt{2} \pm \sqrt{60-42 \sqrt{2}} \text {. }
$$
(Figure 2 takes the negative sign, Figure 3 takes the positive sign)