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Math Help - Analytic Geometry

  1. #1
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    Analytic Geometry

    Find the equation of the circle touching the x-axis and passing through the points (7,-2) and (0,-9).

    Please help me with this and show the solution... Thanks!
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  2. #2
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    Quote Originally Posted by lance View Post
    Find the equation of the circle touching the x-axis and passing through the points (7,-2) and (0,-9).

    Please help me with this and show the solution... Thanks!
    from the given points it can be seen that the circle touches both the axes. also the circle lies in the fourth quadrant.now since the axes are tangents the centre of the circle lies on y=-x. so the centre is (9,-9). you are given 2 points. so calculate the radius. then using circles equation get the required equation.
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  3. #3
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    Hello, lance!

    This is a messy one . . .


    Find the equation of the circle touching the x-axis
    and passing through the points A(7,-2) and B(0,-9).
    Code:
           |
           |        T 
        ---+------*-o-*------------
           | Ao   (h,0)   *
           |*               *
          Bo                 *
           |
          *|        C         *
          *|        *         *
          *|      (h,k)       *
           | 
           *                 *
           |*               *
           |  *           *
           |      * * *
           |

    \text{The center is }C(h,k)\text{, directly below the point of tangency, }T(h,0)

    CA, CB\text{ and }CT\text{ are radii. }\;\text{Hence: }\,CA \,=\,CB \,=\,k


    We have: . \begin{array}{ccccc}<br />
CA^2 & =& (h-7)^2 + (k+2)^2 &=& k^2\\<br />
CB^2 &=& h^2 + (k+9)^2 &=& k^2\end{array}


    Simplify: . \begin{array}{ccccc}<br />
h^2 - 14h + 4k + 53 &=& 0 & {\color{blue}[1]} \\<br />
h^2 \qquad + 18k + 81 &=& 0 & {\color{blue}[2]} \end{array}


    Subtract [2] - [1]: . 14h + 14k + 28 \:=\:0 \quad\Rightarrow\quad k \:=\:-(h+2)\;\;{\color{blue}[3]}


    Substitute into [2]: . h^2 - 18(h+2) + 81 \:=\:0 \quad\Rightarrow\quad h^2 - 18h + 45 \:=\:0

    Hence: . (h-3)(h-15) \:=\:0 \quad\Rightarrow\quad h \:=\:3,\:15

    Substitute into [3]: . k \:=\:-5,\:-17


    We have two solutions: . \begin{array}{|c|c|ccc|} \hline<br />
\text{Center} & \text{Radius} & \text{Equation} &&\\ \hline \\[-5mm]<br />
(3,-5) & 5 & (x-3)^2 + (y+5)^2 &=& 25 \\ \\[-5mm]<br />
(15,-17) & 17 & (x-15)^2 + (y+17)^2 &=& 289 \\ \hline\end{array}

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  4. #4
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    Hello!

    Thank you very much Soroban... It really help me understand the problem... And also to you Pulock2009...
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