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Math Help - A circles problem

  1. #1
    Newbie PowerInside's Avatar
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    A circles problem

    If there are two circles such that (x-a)^2 + (y-b)^2 &= c^2 and (x-b)^2 + (y-a)^2 &= c^2 touch each other,

    Then what can I write for " a &= ? "
    Last edited by PowerInside; August 8th 2010 at 09:34 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    First You with simple algebraic steps write each equation in the form...

    \displaystyle \alpha (x,y,b)\ a^{2} + \beta (x,y,b)\ a + \chi (x,y,b) = 0 (1)

    ... and then find a with the well known quadratic equation resolution formula...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by PowerInside View Post
    If there are two circles such that (x-a)^2 + (y-b)^2 &= c^2 and (x-b)^2 + (y-a)^2 &= c^2 touch each other,

    Then what can I write for " a &= ? "
    The two circles have the following properties:

    - the circles are congruent
    - common tangent of both circles is the line y = x
    - each circle is the image of the other circle by reflection over the line y = x
    - the distance between the centers equals the diameter of one circle

    d = \sqrt{(a-b)^2+(b-a)^2}=2c

    Therefore:

    \sqrt{2(a-b)^2}=2c~\implies~2(a-b)^2=4c^2

    Solve for a: a = b+c\sqrt{2}
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  4. #4
    Newbie PowerInside's Avatar
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    Shouldnt the final answer be  a &= b \pm c\sqrt{2}
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  5. #5
    Newbie PowerInside's Avatar
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    Quote Originally Posted by chisigma View Post
    First You with simple algebraic steps write each equation in the form...

    \displaystyle \alpha (x,y,b)\ a^{2} + \beta (x,y,b)\ a + \chi (x,y,b) = 0 (1)

    ... and then find a with the well known quadratic equation resolution formula...

    Kind regards

    \chi \sigma
    I don't understand this approach. Is it relevent for solving this problem? I did not understand the 3 variable equation you gave to use? ??

    Sorry for being a newbie , But I have difficulty in understanding this one. And im trying to sharpen my math skills..
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  6. #6
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    Quote Originally Posted by PowerInside View Post
    Shouldnt the final answer be  a &= b \pm c\sqrt{2}
    You are right. Value of a will be different depending on the circles touching internally or externally.
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  7. #7
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    Quote Originally Posted by sa-ri-ga-ma View Post
    You are right. Value of a will be different depending on the circles touching internally or externally.
    I don't want to pick at you, but:
    If there are two circles such that (x-a)^2 + (y-b)^2 &= c^2 and (x-b)^2 + (y-a)^2 &= c^2 touch each other,

    Then what can I write for " a = ? "
    According to the text of the question both circles have the same radius. I tried to find two congruent circles where one circle touches the other internally. To be honest I wasn't very successful.

    To be exact you have to consider 3 cases:

    1. a > b, that means the circle with M(a, b) is placed below the line y = x

    2. a = b, that means both circles have the same midpoint on the line y = x (and only in this case there could be an internal touching)

    3. a < b, that means the circle with M(a, b) is placed above the line y = x

    I've used only the first case because the 3rd case is the reflection of the 1st case over the line y = x.
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  8. #8
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    Quote Originally Posted by earboth View Post
    I don't want to pick at you, but:


    According to the text of the question both circles have the same radius. I tried to find two congruent circles where one circle touches the other internally. To be honest I wasn't very successful.
    .
    Yes. You are right.
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  9. #9
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    As earboth shows a= b +crad2 when a is greater than b, a=b -crad 2 when b is greater than a . Another point y=x and y=-xand the two tangent circles can be in any of the 4 quadrants.


    bjh
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