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Math Help - Maths exam tomorrow need help with a circle proof please

  1. #1
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    Maths exam tomorrow need help with a circle proof please

    C1: x^2 + y^2 + 2gx +2fy +c=0
    C2: x^2 + y^2 + 2g'x +2f'y +c'=0

    p and q are the centers of C1 and C2 respectively, show that 2gg'+ 2ff' = c+ c'

    please is someone could help that would be amazing thank you!
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  2. #2
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    Quote Originally Posted by Ilusa View Post
    C1: x^2 + y^2 + 2gx +2fy +c=0
    C2: x^2 + y^2 + 2g'x +2f'y +c'=0
    p and q are the centers of C1 and C2 respectively, show that 2gg'+ 2ff' = c+ c'
    As written, I do not understand the question.
    There must be more to it than you have given us.
    There must be a mistake in "p and q are the centers of C1 and C2 respectively".
    What more can you tell us about it?
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  3. #3
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    Oh my goodness I'm so sorry.

    there is meant to be more

    C1: x^2 + y^2 + 2gx +2fy +c=0
    C2: x^2 + y^2 + 2g'x +2f'y +c'=0

    are two circles

    p and q are the centers of C1 and C2 respectively, and t is the point of intersection of the two circles.
    given that pt is perpendicular to qt, show that show that 2gg'+ 2ff' = c+ c'
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  4. #4
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    Quote Originally Posted by Ilusa View Post
    C1: x^2 + y^2 + 2gx +2fy +c=0 & C2: x^2 + y^2 + 2g'x +2f'y +c'=0 are two circles
    p and q are the centers of C1 and C2 respectively, and t is the point of intersection of the two circles. given that pt is perpendicular to qt, show that show that 2gg'+ 2ff' = c+ c'
    Let \left( {r_p } \right)\,\& \,\left( {r_q } \right) be the radii of the circles centered at p & q respectively.
    By completing squares we get \left( {r_p } \right)^2  = g^2  + f^2  - c\;\,\& \,\,\left( {r_q } \right)^2 = \left( {g'} \right)^2  + \left( {f'} \right)^2  - c'.
    From the given \left[ {l\left( {\overline {pg} } \right)} \right]^2  = \left( {r_p } \right)^2  + \left( {r_q } \right)^2
    What points are P & q?
    Can you finish?
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  5. #5
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    sorry I dont' get this bit
    <br />
\left[ {l\left( {\overline {pg} } \right)} \right]^2<br />

    the points p and q are not given but looking at the diagram the radii of the two circles are equal.
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  6. #6
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    Quote Originally Posted by Ilusa View Post
    sorry I dont' get this bit
    <br />
\left[ {l\left( {\overline {pg} } \right)} \right]^2<br />
the points p and q are not given but looking at the diagram the radii of the two circles are equal.
    \left[ {l\left( {\overline {pg} } \right)} \right] is the distance from p to q.
    P=(-g,-f) & Q=(-g',-f') so what is that distance?
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  7. #7
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    oh thank you, well I wrote that distance as (r^2+r'^2)^1/2
    as distance pt=radius of C1
    and distance qt=raius of C2

    I drew a right angled triangle and used pytagoras' therom to get distance pq

    but I still can't prove it
    Last edited by Ilusa; February 19th 2008 at 02:37 PM. Reason: mad an error in my calculation
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  8. #8
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    \left[ {l\left( {\overline {pg} } \right)} \right]^2  = \left( {g' - g} \right)^2  + \left( {f' - f} \right)^2  = \left( {g^2  + f^2  - c} \right) + \left( {\left( {g'} \right)^2  + \left( {f'} \right)^2  - c'} \right)
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  9. #9
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    thank you so much, you're so kind and intelligent! =)

    sorry to bother you but where do you get (g'-g)^2 + (f'-f)^2 from?
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  10. #10
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    oh sorry it was the distance formula squared sorry! Thank you so much again!
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  11. #11
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    Quote Originally Posted by Ilusa View Post
    where do you get (g'-g)^2 + (f'-f)^2 from?
    Do you know the formula for the distance between two points?
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  12. #12
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    yeah I do thanks, I understand it fully now thank you again!
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