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Math Help - Implicit differentiation to find special points on a circle

  1. #1
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    Implicit differentiation to find special points on a circle

    This was on last years paper for my upcoming test. My workings got so convoluted that I suspect I have strayed off track. I was wondering if I have overlooked something simple and made it more complicated than it needs to be?

    Use implicit differentiation to find the points on the circle x^2-2y+y^2-2x-2=0 closest to and farthest away from the origin.

    Taking the derivative I get:

    \frac{dy}{dx} = \frac{2-2x}{2y-x} = \frac{1-x}{y-1}

    Setting the derivative = 0, I get x = 1 and therefore the top and bottom of the circle: (1,3) and (1,-1)

    After graphing the function, I can see that the two points will be where the line y=x intersects the circle.

    To find where this happens, I tried to find where the gradient of the circle is perpendicular to that of the line y=x.

    \frac{1-x}{y-1} = -1

    1-x=1-y

    y=x :S

    So then I tried setting the original equation = y-x

    x^2-2y+y^2-2x-2 = y-x

    y^2-3y=x-x^2+2

    y^2-3y+\frac{9}{4}=x-x^2+\frac{17}{4}

    (y-\frac{3}{2})^2=x-x^2+\frac{17}{4}

    y = \sqrt{x-x^2+\frac{17}{4}}+\frac{3}{2}

    I'm lost on this one :S
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  2. #2
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    Re: Implicit differentiation to find special points on a circle

    Never mind, after reading my own post I think I have it. I noticed that since the points occur on the line y=x, then the x and y values will equal each other. So, I just re-wrote the original formula for the circle substituting x = y into it to get:

    x^2-2x+x^2-2x-2=0

    x^2-2x-1=0

    x= \frac{2\pm \sqrt{8}}{2} = 1\pm \sqrt{2}

    Which looks right. But I didn't really use implicit differentiation to find the two points did I? It was more geometry :S is there another way?

    I hope I don't get that on my test
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