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Math Help - Circle theorems

  1. #1
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    Circle theorems

    Hi, could anybody help me with this?
    the question:
    what is the distance from the centre (O) to C



    if you need any clarification just ask
    Any help appreciated!
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  2. #2
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    Looking at the modified drawing, recall the secant theorem.
    (AC)(BC)=(EC)(DC). Then solve for x.
    Attached Thumbnails Attached Thumbnails Circle theorems-21870685kw9.gif  
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  3. #3
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    Hello, Geometor!

    I have a solution . . . hope it's acceptable.


    What is the distance from the centre (O) to C?

    Let x \:=\:OC.

    Draw an altitude from O to AB; call it OD.

    In right triangle ODA,\:OA = 5,\:AD = 4\quad\Rightarrow\quad \cos A \:=\:\frac{4}{5}


    Law of Cosines: . OC^2 \;=\;AC^2 + OA^2 - 2(AC)(OA)\cos A

    So we have: . x^2 \;=\;18^2 + 5^2 - 2(18)(5)\left(\frac{4}{5}\right) \;=\;205

    Therefore: . x\;=\;\sqrt{205}

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  4. #4
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    thank you plato!
    and soroban!
    however i find using plato's secant theorem being the easier:
    since:
    (AC)(BC)=(EC)(DC).
    (18)(10)=(10+x)(x)
    =x^2+10x
    x^2+10x-180=0
    (-10 +/- sq.root 82) / 2
    and since we need a positive value we do:
    (-10 + sq.root 82) / 2
    = 9.317821063....
    and add the 5cm radius to get OC
    = 14.3cm(1dp)
    =sq. root 205 as soroban said

    Thanks for the help!
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Geometor!

    I have a solution . . . hope it's acceptable.


    Let x \:=\:OC.

    Draw an altitude from O to AB; call it OD.

    In right triangle ODA,\:OA = 5,\:AD = 4\quad\Rightarrow\quad \cos A \:=\:\frac{4}{5}


    Law of Cosines: . OC^2 \;=\;AC^2 + OA^2 - 2(AC)(OA)\cos A

    So we have: . x^2 \;=\;18^2 + 5^2 - 2(18)(5)\left(\frac{4}{5}\right) \;=\;205

    Therefore: . x\;=\;\sqrt{205}

    That is also my method except that I didnít use trigonometry. By Pythagorasí theorem on triangle ODA, OD = \sqrt{5^2-4^2} = 3 cm. By Pythagorasí theorem on triangle ODC, OC = \sqrt{(4+10)^2+3^2}=\sqrt{205} cm.

    Quote Originally Posted by Geometor View Post
    thank you plato!
    and soroban!
    however i find using plato's secant theorem being the easier:
    since:
    (AC)(BC)=(EC)(DC).
    (18)(10)=(10+x)(x)
    =x^2+10x
    x^2+10x-180=0
    (-10 +/- sq.root 82) / 2
    and since we need a positive value we do:
    (-10 + sq.root 82) / 2
    = 9.317821063....
    and add the 5cm radius to get OC
    = 14.3cm(1dp)
    =sq. root 205 as soroban said

    Thanks for the help!
    You made a mistake there; it should be 820, not 82. Itís also easier to make mistakes with your calculations using the secant method. My recommendation: use the method that Soroban and I used.
    Last edited by JaneBennet; December 21st 2007 at 11:20 AM.
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  6. #6
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    haha thanks for pointing it out
    i wrote 820 in my calculations though phew
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, Geometor!

    I have a solution . . . hope it's acceptable.

    Let x \:=\:OC.

    Draw an altitude from O to AB; call it OD.

    In right triangle ODA,\:OA = 5,\:AD = 4\quad\Rightarrow\quad \cos A \:=\:\frac{4}{5}


    Law of Cosines: . OC^2 \;=\;AC^2 + OA^2 - 2(AC)(OA)\cos A

    So we have: . x^2 \;=\;18^2 + 5^2 - 2(18)(5)\left(\frac{4}{5}\right) \;=\;205

    Therefore: . x\;=\;\sqrt{205}
    How do you know AD=4?
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  8. #8
    Senior Member JaneBennet's Avatar
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    Because OAB is an isosceles triangle (so the perpendicular from O to AB bisects AB).
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  9. #9
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    Oh right, sorry it's 2:07 AM here lol
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  10. #10
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    Quote Originally Posted by JaneBennet View Post
    My recommendation: use the method that Soroban and I used.
    Reading the title of the posting “Circle Theorems” why would you recommend against using a very basic theorem about circles? The approach that you advocate is not unique to theorems about circles but rather belongs to the general discussion about triangles.
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  11. #11
    Senior Member JaneBennet's Avatar
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    I didnít realize you had to use the theorem to solve this problem. My apologies.
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