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Math Help - tangent, circle

  1. #1
    Junior Member
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    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
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    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 L has equation y\:=\:mx
    and a circle C has equation x^2+y^2-6x-4y+9\:=\:0

    Given that L is tangent to C, to find the possible values of m.
    The line y \:=\:mx contains the origin.

    The circle has the equation: . (x-3)^2+(y-2)^2\:=\:4
    Its center is C(3,\,2) and it has radius 2.
    It is tangent to the x-axis at A(3,0).
    Code:
      |
      |                     * * *
      |                 *           *
      |               *               *
      |            B o                 *
      |                 * 2
      |             *      *            *
      |         *   *         oC        *
      |             *     *   :         *
      |               *       :
      |    * θ    *  *        :2       *
      |       *       *       :       *
      |   *  θ          *     :     *
    - o . - - - - - - - - - * o * - - - - - -
      O           3           A
    Obviously, one tangent is: . y \,= \,0\quad\Rightarrow\quad\boxed{m \:=\:0}

    Let \theta = \angle COA
    The slope of OC is: . \tan\theta \:=\:\frac{2}{3}

    From congruent right triangles, \angle COB = \theta
    The slope of OB is: . \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}}

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