The line $a x+b y-12=0$ passes through the center of the circle $(0,2) \Rightarrow b=6$. Let $S(t,-6), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $S A: x_{1} x+\left(y_{1}-2\right)(y-2)=16, S B: x_{2} x+\left(y_{2}-2\right)(y-2)=16$, substituting $S(t,-6)$ yields $\left\{\begin{array}{l}x_{1} t-8 y_{1}=0, \\ x_{2} t-8 y_{2}=0\end{array} \Rightarrow A B: t x-8 y=0\right.$, so the line $A B$ passes through the fixed point $(0,0)$.
4. Circle $C: x^{2}+(y-2)^{2}=16$ is symmetric about the line $a x+b y-12=0$, a moving point $S$ is on the line $y+b=0$, two tangent lines $S A, S B$ are drawn from point $S$ to circle $C$, with points of tangency $A, B$ respectively, then the line $A B$ must pass through a fixed point, the coordinates of the fixed point are $\qquad$