Answer
$$y=\pm \frac{\sqrt{b^2-9}}{3}+b$$
Work Step by Step
The general equation of a line is $y=mx+b$, and the equation of the circle with center at the origin and radius $3$ is $x^2+y^2=9$.
If we want that the line is tangent to the circle, we must have $$x^2+(mx+b)^2=9 \quad \Rightarrow \quad (1+m^2)x^2+2mbx+b^2-9=0 \\ \Delta =(2mb)^2-4(1+m^2)(b^2-9)=0 \quad \Rightarrow \quad b^2-9-9m^2=0 \quad \Rightarrow \quad m=\pm \frac{\sqrt{b^2-9}}{3}.$$Thus, an equation for the family of lines tangent to the circle with center at the origin and radius $3$ is $$y=\pm \frac{\sqrt{b^2-9}}{3}+b \, .$$