4. The vertex $C$ of $\triangle A B C$ has coordinates $(x, y)$ that satisfy the inequalities $x^{2}+y^{2} \leqslant 8+2 y, y \geqslant 3$. The side $A B$ lies on the x-axis. If it is known that the point $Q(0,1)$ is at a distance of 1 from the lines $A C$ and $B C$, find the maximum area of $\triangle A B C$.
From the given information, point $C$ lies on the circle with center $Q$ and a radius of 3, within the shaded area, i.e., the horseshoe shape $K L M N$ (as shown in the figure). The circle $S$ with center $Q$ and a radius of 1 is the incircle of $\triangle A B C$. The equation of the tangent line to $\odot S$ at point $C$ can be written as:

Let $a$ be the x-coordinate of the intersection point of this tangent line with the $O X$ axis, and $(X, Y)$ be the coordinates of any point on the tangent line. Then the equation of the tangent line is:
$$
{ }_{-X-a}^{x-a}=\frac{Y}{y} \text { or } Y=\frac{y}{x-a} X-\frac{a y}{x-a} .
$$

The equation of circle $S$ is:
$$
X^{2}+(Y-1)^{2}=1 \text {. }
$$

The line given by equation (1) is a tangent to circle (2), which means the system of equations (1) and (2) has a unique solution. By eliminating $Y$ from this system, we get the equation:
$$
X^{2}+\left[\frac{y}{x-a} X-\left(\frac{a}{x-a}+1\right)\right]^{2}=1,
$$

which has a unique solution. In this case, the discriminant $\Delta=0$ of this quadratic equation.
To determine $a$, we transform the equation, yielding:
$(y-2) a^{2}+2 x a-y=0$, whose roots $a_{1}$ and $a_{2}$ are the x-coordinates of the vertices $A$ and $B$ of $\triangle A B C$.
Since $|A B|=\left|a_{1}-a_{2}\right|$
$$
=\frac{2}{y-2} \sqrt{x^{2}+y(y-2)} \text {, }
$$

the length of segment $A B$ is maximized when $x^{2}$ is maximized for a fixed value of $y(3 \leqslant y \leqslant 4)$. For a fixed $y$ value, the maximum value of $x^{2}$ is $9-(y-1)^{2}$. Therefore, when $y \in[3, 4]$, the maximum value of $A B$ is:
$$
\begin{array}{l}
\frac{2}{y-2} \sqrt{9-(y-1)^{2}+y(y-2)} \\
=\frac{4 \sqrt{2}}{y-2} .
\end{array}
$$

This value is achieved when $C$ coincides with $C_{1}$ or $C_{2}$. Here, $C_{1}, C_{2}$ are points on the arc $\overparen{K L M}$ with the same y-coordinate $y$. The maximum area of $\triangle A B C$ is:
$$
\begin{aligned}
& S(y)=S_{\triangle A B C}=\frac{1}{2} \cdot \frac{4 \sqrt{2}}{y-2} \cdot y \\
= & 2 \sqrt{2} \cdot\left(1+\frac{2}{y-2}\right) .
\end{aligned}
$$

Thus, it does not exceed $6 \sqrt{2}$, and the equality holds when $y=3$, i.e., when vertex $C$ coincides with one of the endpoints of the chord $K M$ of the horseshoe shape $K L M N$. At this time, the maximum value of $\triangle A B C$ is $6 \sqrt{2}$.

Note: The minimum area of $\triangle A B C$ is $3 \sqrt{3}$, which is achieved when $C$ coincides with $N$.