4. For all real number pairs $(x, y)$ satisfying $(x-3)^{2}+(y-3)^{2}=6$, the maximum value of $\frac{y}{x}$ is ( ).
(A) $3+2 \sqrt{2}$
(B) $4+\sqrt{2}$
(C) $5+3 \sqrt{3}$
(D) $5+\sqrt{3}$
4. (A).

Let $\frac{y}{x}=t$, then $\left(t^{2}+1\right) x^{2}-6(t+1) x+12=0$.
We have $\Delta=36(t+1)^{2}-48\left(t^{2}+1\right) \geqslant 0, t^{2}-6 t+1 \leqslant 0$.
Since $t^{2}-6 t+1=[t-(3-2 \sqrt{2})][t-(3+2 \sqrt{2})]$,
we know that the solution set of $t^{2}-6 t+1 \leqslant 0$ is $3-2 \sqrt{2} \leqslant t \leqslant 3+2 \sqrt{2}$.
Therefore, $\left(\frac{y}{x}\right)_{\max }=3+2 \sqrt{2}$.