1. ## tangent, circle

a line l has equation y=mx and a circle C has equation x^2+y^2-6x-4y+9=0

Given that l is a tangent to C, I have to find the possible values of m.

How would i go about doing this

Thanks

2. 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 $\displaystyle x^2+y^2-6x-4y+9\:=\:0$

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: .$\displaystyle (x-3)^2+(y-2)^2\:=\:4$
Its center is $\displaystyle C(3,\,2)$ and it has radius 2.
It is tangent to the x-axis 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: .$\displaystyle \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{2\left(\frac{2}{3}\right)}{1 - \left(\frac{2}{3}\right)^2}\quad\Rightarrow\quad\b oxed{m \;=\;\frac{12}{5}}$