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Thread: intersecting circles

  1. #1
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    intersecting circles

    Hi folks,

    I am trying to find the intersection points A and B of the two circles shown in the attachment. I am getting imaginary roots, which I was not expecting.

    The equations of the circles are:

    $x^2 + y^2 - 4x = 0$ and

    $x^2 + y^2 -4y + 3 = 0$

    subtracting one from the other gives $-4x + 4y - 3 = 0 \Rightarrow y = x + \frac{3}{4}$

    substituting in the first circle: $ x^2 + (x + \frac{3}{4})^2 - 4x = 0$

    $2x^2 + \frac{3}{2}x + \frac{9}{4} - 4x = 0 \Rightarrow 2x^2 - \frac{5}{2}x + \frac{9}{4} = 0$

    $8x^2 - 10x + 9 = 0$

    a quadratic in x with a = 8, b = -10 and c = 9 produces a discriminant $(b^2 - 4ac) < 0$ i.e. no real roots.

    Clearly I am making a mistake somewhere! Can anyone see it?
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    Last edited by s_ingram; Jul 13th 2017 at 06:43 AM.
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  2. #2
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    Re: intersecting circles

    Thanks from s_ingram
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  3. #3
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    Re: intersecting circles

    $\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16} \neq \dfrac{9}{4}$
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  4. #4
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    Re: intersecting circles

    many thanks.
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  5. #5
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    Re: intersecting circles

    Quote Originally Posted by SlipEternal View Post
    $\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16} \neq \dfrac{9}{4}$
    Sir Ingram, go stand in the corner for 9/16 hour.....
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  6. #6
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    Re: intersecting circles

    You're right! I am deeply ashamed.
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