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Thread: Circle problem

  1. #1
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    Circle problem

    I know what happens because i played around with software, but i cant see an easy way why?
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  2. #2
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    Re: Circle problem

    coordinates:

    A\left(a,\sqrt{a(2-a)}\right)
    and
    B\left(0,\sqrt{2a}\right)

    where 0<a<1

    write down the equation of line AB and find its intersection with the x-axis

    C\left(2+\sqrt{4-2a},0\right)

    take the limit as a\to  0
    Last edited by Idea; Nov 26th 2017 at 11:12 AM.
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  3. #3
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    Re: Circle problem

    why is the y coordinate of B sqrt(2a) ?

    Ok ignore ..use the property that OA=OB ..

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  4. #4
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    Re: Circle problem

    Now, I cant see how you got a neat form of the x coord of C! I get a very messy expression..

    Ok ..so you rationalised a surd. Is there a Geometric way of looking at this? If I change the circle the x coord of C appears to be double the diameter.
    Last edited by rodders; Nov 26th 2017 at 01:00 PM.
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  5. #5
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    Re: Circle problem

    If AH is perpendicular to OC, H on OC, we can use similar triangles to get

    \frac{\text{CH}}{\text{CO}}=\frac{\text{AH}}{\text  {BO}}

    \frac{x-a}{x}=\frac{\sqrt{a(2-a)}}{\sqrt{2a}}

    Solve for x

    x=\frac{2 a}{2 - \sqrt{4-2a }}

    rationalize
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