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Math Help - Finding radius of circle with 2 Perpendicular Chords

  1. #1
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    Finding radius of circle with 2 Perpendicular Chords

    Two chords KV and QR of circle O are perpendicular at point P, with PQ=6 and PR=8 If radius of circle is sq root of 65. find KP and PV.


    The answer is KP=12 and PV=4

    My ques is how did they get KP? i understand how they get PV since QP*PR=KP*PV

    But i dont know what to do with the radius to get KP?
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  2. #2
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    Re: Finding radius of circle with 2 Perpendicular Chords

    Quote Originally Posted by Aquameatwad View Post
    Two chords KV and QR of circle O are perpendicular at point P, with PQ=6 and PR=8 If radius of circle is sq root of 65. find KP and PV.


    The answer is KP=12 and PV=4

    My ques is how did they get KP? i understand how they get PV since QP*PR=KP*PV

    But i dont know what to do with the radius to get KP?
    1. Draw a sketch.

    2. Let x denote |\overline{VP}| and y denotes |\overline{PK}|.
    According to the intersecting chord theorem (google for it!)[1st equation] and Pythagorean theorem[2nd equation] you'll get the system of equations:

    \left|\begin{array}{rcl}x \cdot y &=& 6 \cdot 8 \\ \left(\dfrac{x+y}{2}\right)^2+1^2 &=& 65 \end{array}\right.

    3. Solve for x and y.
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    Re: Finding radius of circle with 2 Perpendicular Chords

    I understand the intersecting chord theorem. But how did you derive the 2nd equation?
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    Re: Finding radius of circle with 2 Perpendicular Chords

    Quote Originally Posted by Aquameatwad View Post
    I understand the intersecting chord theorem. But how did you derive the 2nd equation?
    The centre lies on the perpendicular bisector of QR,
    which is parallel to KV
    and lies 1 unit below KV (from the sketch).
    Therefore Pythagoras' Theorem applied to the right-angled triangle OKM or OVM,
    where O is the centre and M is the midpoint of KV yields Earboth's 2nd equation.
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    Re: Finding radius of circle with 2 Perpendicular Chords

    Thanks. i get it now. i was making it way to complicated
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  6. #6
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    Re: Finding radius of circle with 2 Perpendicular Chords

    You could also have answered exclusively with Pythagoras' Theorem.

    Label the circle centre O, the midpoint of KV = M and the midpoint of QR = N.

    Then the circle centreline containing O and N lies 1 unit below [KV]
    as it is 7 units from Q and 7 units from R.

    M is the midpoint of [KV]

    \Rightarrow\ |KM|^2+1=65\Rightarrow\ |KM|=8

    Label the point of intersection of [QR] and the horizontal centreline as S.

    Then

    |OS|^2+7^2=65\Rightarrow\ |OS|=4

    Then

    |KV|=2|KM|,\;\;\;|KP|=|KM|+|OS|,\;\;\;|PV|=|KM|-|OS|
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