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Math Help - Equation for a circle.

  1. #1
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    Equation for a circle.

    Hello.

    I want to rewrite this equation with respect to y

    x^2 + (y-3) = 25

     y^2 - 6y = 25-9 - x^2

     y^2 - 6y = 16-x^2

    Please help
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  2. #2
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    Quote Originally Posted by MatteNoob View Post
    Hello.

    I want to rewrite this equation with respect to y

    x^2 + {\color{red}(y-3)^2} = 25

     y^2 - 6y = 25-9 - x^2

     y^2 - 6y = 16-x^2

    Please help
    Note the correction in red.
    x^2 + {\color{red}(y-3)^2} = 25

    \Rightarrow (y-3)^2 = 25 - x^2

    \Rightarrow y - 3 = \pm \sqrt{25 - x^2}

    \Rightarrow y = \pm \sqrt{25 - x^2} + 3
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  3. #3
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    Mr Fantastic

    Oh, sorry about the typo.

    So if I want to write the equation with respect to y, I could do this:

    (x-3)^2 +(y+1)^2 = 25

    (y+1)^2 = 5^2 - (x-3)^2

    y = \pm \sqrt{5^2 - (x-3)^2}-1

    Then, if I want the derivative, I could;

    u = -x^2 + 6x -16 \;\;\;\; u\prime = -2x +6

    y\prime = \pm \left(\sqrt{u}\right)\prime \cdot u\prime - (1)\prime

    So that

     y\prime = \pm \frac{-2x+6}{2\sqrt{-x^2+6x-16}} = \underline{\underline{ \pm \frac{3-x}{\sqrt{6x - x^2-16}}}}

    Edit: Corrected a typo.
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  4. #4
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    Quote Originally Posted by MatteNoob View Post
    Mr Fantastic

    Oh, sorry about the typo.

    So if I want to write the equation with respect to y, I could do this:

    (x-3)^2 +(y+1)^2 = 25

    (y+1)^2 = 5^2 - (x-3)^2

    y = \pm \sqrt{5^2 - (x-3)^2}-1

    Then, if I want the derivative, I could;

    u = -x^2 + 6x {\color{red}+} 16 \;\;\;\; u\prime = -2x +6

    y\prime = \pm \left(\sqrt{u}\right)\prime \cdot u\prime - (1)\prime

    So that

     y\prime = \pm \frac{-2x+6}{2\sqrt{-x^2+6x {\color{red}+} 16}} = \underline{\underline{ \pm \frac{3-x}{\sqrt{6x - x^2 {\color{red}+} 16}}}}

    Edit: Corrected a typo.
    Note the corrections in red.
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