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Math Help - Circle Geometry

  1. #1
    Newbie Femto's Avatar
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    Circle Geometry

    The point A has coordinates (2, -1). Find the equation of the chord of the circle which has A as its midpoint.

    I don't know how to do it with limited information. Could somebody help please? I'm a bit shoddy at these questions.

    It has preceding questions which say let C be the circle with equation x^2 + y^2 - 2x + 4y - 11 = 0

    I found the centre and radius.

    Then it asked me to determine whether the origin lies inside or outside C; it lies inside.
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  2. #2
    MHF Contributor

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    As written there are infinitely many answers.
    Therefore, you have posted only a very small part of the complete question.
    Please revise the question giving all the information.
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  3. #3
    Newbie Femto's Avatar
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    Well the centre of C is (1, -2) and the radius is 4.

    I found that the origin lies inside the circle since the distance to the origin from the centre is only \sqrt 5 which is smaller than 4.

    They are the only preceding pieces of information I have.
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  4. #4
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    The slope of the line determined by A~\&~C is \dfrac{(-2)-(-1)}{(1)-(2)}=1.
    Therefore the slope of the chord is -1. WHY?
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  5. #5
    Newbie Femto's Avatar
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    Quote Originally Posted by Plato View Post
    The slope of the line determined by A~\&~C is \dfrac{(-2)-(-1)}{(1)-(2)}=1.
    Therefore the slope of the chord is -1. WHY?
    Since the chord is perpendicular to the tangent gradient you've found?

    But how is AC a tangent gradient?
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  6. #6
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    Quote Originally Posted by Femto View Post
    Since the chord is perpendicular to the tangent gradient you've found?
    But how is AC a tangent gradient?
    The line determined by the center of a circle and the midpoint of a chord is the perpendicular bisector of the chord.
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  7. #7
    Newbie Femto's Avatar
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    Oh right thanks very much I forgot about that!
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