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Math Help - question on properties of circle 8

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    question on properties of circle 8

    Referring to the attached pic:
    APQ is a semi-circle with PQ as the diameter.
    QP is extended to B, and BA is a tangent to the circle at A.
    AC is perpendicular to PQ and PC:CQ = 1:4.
    AB = 8cm
    (a) Prove AC^2 = (PC)(QC)
    (b) Prove AB^2 = (BP)(BQ)
    (c) If BP = y and PC = k, prove that k = \frac{3}{5} y
    (d) Find BP
    (e) Find the radius of circle.

    My work:
    (a)
    triangles APC ~ ACQ (same angles)
    hence AC:PC = QC:AC
    AC^2 = (PC)(QC)
    (b)
    triangles BAQ ~ BPA (same angles)
    hence AB:BQ = BP:AB
    AB^2 = (BP)(BQ)
    (c)
    since PC:CQ = 1:4,
    CQ = 4k and hence PQ = 5k
    hence \frac{k}{AC} = \frac{AC}{4k}
    AC^2 = 4k^2 = (2k)^2
    hence we have AC = 2k
    AP^2 = AC^2 + PC^2 (Pythagarus)
    AP^2 = (2k)^2 + k^2
    AP^2 = 5k^2
    AP = (k)(5^0.5)
    Also AQ^2 = PQ^2 - AP^2
    AQ^2 = (5k)^2 - 5k^2
    AQ^2 = 20k^2
    AQ = (2k)(5^0.5)
    y:8 = AP:AQ = k(5^0.5):2k(5^0.5)
    y:8 = 1:2
    hence y = 4
    and 8:BQ = AP:AQ = 1:2
    8:(5k + y) = 1:2
    8:(5k + 4) = 1:2
    16 = 5k + 4
    k = \frac{12}{5}
    Therefore, \frac{3}{5} y = (\frac{3}{5})(4) = \frac{12}{5} = k
    Confused here: relation between k and y is supposed to be proved without working out their values in advance. Is there another method to prove that k = 3y/5 ????
    (d)
    BP = y = 4cm
    (e)
    PQ = 5k = 12cm
    Attached Thumbnails Attached Thumbnails question on properties of circle 8-ques-16.jpg  
    Last edited by ukorov; November 8th 2009 at 07:52 AM.
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