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    I'm nowhere near AEA standard but I was looking at some old papers to see how hard they are and I saw an interesting one.

    http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

    Question 4(a)(i)
    Can anyone post a solution?
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    (Original post by Widowmaker)
    I'm nowhere near AEA standard but I was looking at some old papers to see how hard they are and I saw an interesting one.

    http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

    Question 4(a)(i)
    Can anyone post a solution?
    I get r = 2.

    I think I need a diagram.

    Aitch
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    (0,4) to centre is the same as perpendicular from centre to the tangent.... both are length r
    let the tangential point be called t

    the length from the origin to t is 4 (y axis is also a tangent, length of tangents from the same origin to where they touch the circle is the same)

    equation of line is 4y - 3x = 0, which is also y = (3x)/4

    using pythagoras (4^2) = (x^2) + (y^2)
    (4^2) = (x^2) + (3x/4)^2)
    16 = (x^2) + (9(x^2)/16)
    16 = (x^2) + (9(x^2)/16)
    256 = 16(x^2) + 9(x^2)
    256 = 25(x^2)
    (x^2) = 10.24

    x = 3.2

    if x = 3.2, then y = 2.4...

    gradient of orginal line = 3/4
    gradient of perpendicular = -4/3

    y-y1=m(x-x1)
    y - 2.4 = (-4/3)(x - 3.2)
    y - 2.4 = -4x/3 + 64/15
    y - 20/3 + 4x/3 = 0
    3y + 4x - 20 = 0

    we know the y coordinate of the centre to be 4. if y=4 then...

    12 + 4x - 20 = 0
    4x = 8
    x = 2

    if the centre of the circle has x = 2, and the circumference is at o then the radius, r, = 2
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    (Original post by Aitch)
    I get r = 2.

    I think I need a diagram.

    Aitch
    i had r=2, went to post, got logged out, lost solution, did it again, and got r = 2 again... without diagram...
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    (Original post by El Stevo)
    i had r=2, went to post, got logged out, lost solution, did it again, and got r = 2 again... without diagram...
    :congrats:
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    (Original post by El Stevo)
    i had r=2, went to post, got logged out, lost solution, did it again, and got r = 2 again... without diagram...
    I won't bother with the diagram now!

    I worked with the 2 triangles with smallest angle = arctan 3/4
    Then a perpendicular dropped from the centre meets 4y=3x at 5r/4 below the centre.
    at that point the y coord is 3/4 *r

    so 2r =4 (between centre and x-axis)

    Aitch
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    Don't say you're 'not near AEA standard'. AEA exams are supposed to be difficult and thus you're not expected to get every question right.
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    when is the AEA exam taken? end of AS?
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    (Original post by C4>O7)
    when is the AEA exam taken? end of AS?
    After you finish your A level course.
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    Hey THAT WAS THE PAPER I DID!!!

    Q1.. I left it... but aparently my mate said it was easy... mind you... BE VERY CAREFUL AT THE END.. x=0 DOES NOT SATISFY

    Q2... easy...

    Q3... easy...

    Q4...for the circle... equate the parametric equation of the circle with the cartesian of the line I remember... to get the point of intersection...

    Q5... it was ok...

    Q6 ok...

    Q7... RUN OUT OF TIME...
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    Found my (pretty ropey) diagram!

    Two 3/2,4/2,5/2 triangles.

    a hypotenuse and a shortest side = 4 = 2r !

    Aitch
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