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Math Help - coordinate geometry of circles

  1. #1
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    Question coordinate geometry of circles

    1. (x-3)^2 +(y-2)^2=36
    find equations of the two horizontal tangents to the circle

    2.(x-2)^2 +(y-2)^2 =32
    Find coordinates of the endpoints of the diameter whose equation is y=x

    please respond quickly
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  2. #2
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    [quote=wale;487298]1. (x-3)^2 +(y-2)^2=36
    find equations of the two horizontal tangents to the circle

    2.(x-2)^2 +(y-2)^2 =32
    Find coordinates of the endpoints of the diameter whose equation is y=x

    please respond quickly[/quote

    Need answers fast?
    I will try to help.
    Center of first circle 3,2
    radius of circle is 6

    Plot and id the two tangents.No cals required


    Center of second circle 2,2
    radius of circle is 4radical 2
    Plot the center point and draw in y=x start at origin line has slope = 1 extend line above and below the x axis
    draw in two slope diagrams using the center as reference point but the rise and the run must be 4 so that the hypothenus will be 4 rad 2. No cals required


    bjh
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  3. #3
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    Hello, wale!

    Did you make any sketches?


    1.\;\;(x-3)^2 +(y-2)^2\:=\:36

    Find the equations of the two horizontal tangents to the circle.
    Code:
                  | 
                  | (3,8) 
            - - - + * o * - - -
                * |   :     *
              *   |   :       * 
             *    |   : r=6    * 
                  |   : 
            *     |   :         *
            *     |   o(3,2)    *
            *     |   :         * 
                  |   : 
          ---*----+---:--------*-----
              *   |   :       *
                * |   :     *
            - - - + * o *  - - - 
                  | (3,-4) 
                  |

    Got it?




    2.\;\;(x-2)^2 +(y-2)^2 \:=\:32

    Find the coordinates of the endpoints of the diameter whose equation is y=x
    Code:
                  |
                  | * * * 
                * |         *   *
              *   |           o
             *    |         *  *
                  |       *
            *     |     *       * 
            *     |   o (2,2)   *
            *     | *           *
          - - - - + - - - - - - - - - -
             *  * |            *
              o   |           *
            *   * |         *
                  | * * *
                  |

    We want the intersections of: . \begin{Bmatrix}[1] & (x-2)^2 + (y-2)^2 \:=\:32 \\ [2] & x\:=\:y \end{Bmatrix}


    Substitute [2] into [1]: . (x-2)^2 + (x-2)^2 \:=\:32 \quad\Rightarrow\quad 2(x-2)^2 \:=\:32

    . . . . . . . (x-2)^2 \:=\:16 \quad\Rightarrow\quad x-2 \:=\:\pm4 \quad\Rightarrow\quad x \:=\:2 \pm 4

    Hence: . x \;=\;6,\:-2 \quad\Rightarrow\quad y \;=\;6,\:-2


    Therefore, the endpoints of the diameter are: . (6,6)\,\text{ and }\,(\text{-}2,\text{-}2)

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