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Math Help - Equation of circles help?

  1. #1
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    Equation of circles help?

    The centre of two circles lie on the line x+y-7=0

    Both circles touch the x - axis and contain the point (6,2), find the equations of the two circles that satisfy these conditions.
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  2. #2
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    Hello, sam1234!

    The centres of two circles lie on the line x+y \:=\:7
    Both circles touch the x-axis and contain the point P(6,2).

    Find the equations of the two circles.
    Code:
                |
        *       | * * *
          *   * |         *
            *   |           *
           *  * |            *
                *
          *     | * C         *
          *     |   o (h,k)   *
          *     |   : *       *
                |   :   *
           *    |   :k    *  o P(6,2)
            *   |   :       *
        ------*-+---+-----*---*------
                | * * *         *
                |

    The center is C(h,k), which lies on x+y-7 \:=\:0
    . . Hence: . h+k-7 \:=\:0 \quad\Rightarrow\quad k \:=\:7-h .[1]


    The circle is tangent to the x-axis, so the radius is k.

    The equation is: . (x-h)^2 + (y-k)^2 \:=\:k^2

    Since (6,2) is on the circle: . (6-h)^2 + (2-k)^2 \:=\:k^2  \quad\Rightarrow\quad h^2-12h + 40 - 4k \:=\:0 .[2]


    Substitute [1] into [2]: . h^2-12h + 40 - 4(7-h) \:=\:0 \quad\Rightarrow\quad h^2-8h + 12 \:=\:0

    Then: . (h-2)(h-6) \:=\:0 \quad\Rightarrow\quad h \;=\;2,\:6

    . . Hence: . k \;=\;5,\:1


    Therefore: . \begin{Bmatrix}<br />
C(2,5),\:r=5 && \Rightarrow &&(x-2)^2 + (y-5)^2 &=& 25 \\ \\[-3mm]<br />
C(6,1),\:r=1 && \Rightarrow && (x-6)^2 + (y-1)^2 &=& 1 \end{Bmatrix}

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