Hello, lra11!
Did you make a sketch?
There are a number of approaches to this problem.
Here's one that does not require the points of tangency.
A line $\displaystyle L$ has equation $\displaystyle y\:=\:mx$
and a circle $\displaystyle C$ has equation $\displaystyle x^2+y^26x4y+9\:=\:0$
Given that $\displaystyle L$ is tangent to $\displaystyle C,$ to find the possible values of $\displaystyle m.$ The line $\displaystyle y \:=\:mx$ contains the origin.
The circle has the equation: .$\displaystyle (x3)^2+(y2)^2\:=\:4$
Its center is $\displaystyle C(3,\,2)$ and it has radius 2.
It is tangent to the xaxis at $\displaystyle A(3,0).$ Code:

 * * *
 * *
 * *
 B o *
 * 2
 * * *
 * * oC *
 * * : *
 * :
 * θ * * :2 *
 * * : *
 * θ * : *
 o .          * o *      
O 3 A
Obviously, one tangent is: .$\displaystyle y \,= \,0\quad\Rightarrow\quad\boxed{m \:=\:0}$
Let $\displaystyle \theta = \angle COA$
The slope of $\displaystyle OC$ is: .$\displaystyle \tan\theta \:=\:\frac{2}{3}$
From congruent right triangles, $\displaystyle \angle COB = \theta$
The slope of $\displaystyle OB$ is: .$\displaystyle \tan2\theta \:=\:\frac{2\tan\theta}{1\tan^2\!\theta} \:=\:\frac{2\left(\frac{2}{3}\right)}{1  \left(\frac{2}{3}\right)^2}\quad\Rightarrow\quad\b oxed{m \;=\;\frac{12}{5}}$