1. A ray of light $l$ is emitted from point $A(-3,3)$ and strikes the $x$-axis, where it is reflected. The reflected ray is tangent to the circle $x^{2}+y^{2}-4 x-4 y+7=0$. Find the equation of the line on which the ray $l$ lies.
1. Given the equation of a circle as $(x-2)^{2}+(y-2)^{2}=1$, the equation of its symmetric circle with respect to the $x$-axis is $(x-2)^{2}+(y+2)^{2}=1$. This circle is tangent to a line $l$, and let the point of tangency be $B\left(x_{1}, y_{1}\right)$. For any point $M(x, y)$ on $l$, $B$ divides $M A$ in a fixed ratio $\lambda$, then $x_{1}=\frac{x-3 \lambda}{1+\lambda}, y_{1}=$ $\frac{y+3 \lambda}{1+\lambda}$. Since $B$ lies on the circle, we have $49 \lambda^{2}+2(5 y-5 x+19) \lambda+\left(x^{2}+y^{2}-4 x+4 y+7\right)=0$. Also, since $l$ is tangent to the circle, $\triangle=0$, i.e., $-8(3 x+4 y-3)(4 x+3 y+3)=0$, hence the equations of the two lines are $3 x+4 y-3=0,4 x+3 y+3=0$.