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    Hey,

    am so stuck on this question ... and I thought C1 wasn't that bad! *sigh*

    Anyway, would be cool if someone could help me out!

    **A circle passes through the points A(2,9) and B(10,3). AB is a diameter of the circle.

    (i) Calculate the radius of the circle and the coordinates of the centre. (I don't even know the forumla how to calculate the radius!!)

    (ii) Show that the equation of the circle may be written in the form x^2 + y^2 - 12x -12y + 47 = 0!

    (iii) The tangent to the circle at the point B cuts the x-axis at C. Find the coordinates of C.
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    (i) The radius is half of the diameter. You can calculate the length of the diameter using the co-ordinates given (using Pythagoras).

    d2 = (10-2)2+ (3-9)2
    d2 = 100
    d = 10
    => radius, r = 5

    The centre will be at the midpoint of the diamter.
    x-cord of centre = (2+10)/2 = 6
    y-cord of centre = (9+3)/2 = 6
    => Centre (6,6)


    (ii) Equation circle: (x-6)2 + (y-6)2 = 25
    x2 - 12x + 36 + y2 - 12y + 36 - 25 = 0
    x2 + y2 - 12x - 12y + 47 = 0


    (iii) Differentiate:
    2x + 2y(dy/dx) - 12 - 12(dy/dx) = 0
    (2y-12)(dy/dx) = 12-2x
    dy/dx = (6-x)/(y-6)

    At B, x=10, y=3
    => dy/dx = -4/-3 = 4/3

    Eq. tangent at B:
    y-3 = (4/3)(x-10)

    If this tangent cuts the x-axis at C, then the y-cord of C must be 0. So sub y=0 into the above equation:

    -3 = (4/3)(x-10)
    -9 = 4x - 40
    4x = 31
    x = 31/4

    => C( 31/4 , 0 )
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    Thanks very much ..
 
 
 
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Updated: May 20, 2005

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