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Math Help - Circle tangent to the line.

  1. #1
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    Circle tangent to the line.

    Find the eq. of the circle:
    "The circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

    Can someone help me with this one, I really don't know how to start things up. to be exact, I don't understand the question, sorry.

    Thank you.
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  2. #2
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    Quote Originally Posted by dissidia View Post
    Find the eq. of the circle:
    "The circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

    Can someone help me with this one, I really don't know how to start things up. to be exact, I don't understand the question, sorry.

    Thank you.
    equation of circle centered at (x_{0},0) : (x - x_{0})^2 + y^2 = r^2.
    this circle passed the point (4,2), substitute it, we have

    (4 - x_{0})^2 + 2^2 = r^2

    we also have equation : (x_{0} - 2)^2 = 8 + r ^2 from the tangent line of the circle.

    then

    (4 - x_{0})^2 + 2^2 = (x_{0} - 2)^2 - 8

    16 - 8x_{0} + x_{0}^2 + 2^2 = x_{0}^2 - 4x_{0} + 4 - 8

    16 - 8x_{0} + 2^2 = - 4x_{0} - 4

    4x_{0} = 24

    x_{0} = 6

    and

    6^2 = 20 + r ^2

    r^2 = 8

    the equation of the circle : (x - 6)^2 + y^2 = 8
    Last edited by dedust; January 14th 2010 at 06:03 AM.
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  3. #3
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    Thank you so much for your reply. just a question for me to fully understand it, just where did you get this? [tex]

    (x - x_{0})^2 + y^2 = r^2<br />

    And what happened to x-y=2
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  4. #4
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    ^mate,  (x-a)^2 + (y-b)^2 = r^2
    is the general equation of a circle
    since your circle lies on the x axis
    the centre is  (x_0,0)
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  5. #5
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    Oh yeah I got it right, damn... XD, one last thing, what happened to x-y=2? I mean, what's the use of it? I don't see it used in the equation dedust solved.
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  6. #6
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    Smile

    Quote Originally Posted by dissidia View Post
    Oh yeah I got it right, damn... XD, one last thing, what happened to x-y=2? I mean, what's the use of it? I don't see it used in the equation dedust solved.
    we use it to get this equation


    we also have equation : (x_{0} - 2)^2 = 8 + r ^2 from the tangent line of the circle.
    Last edited by dedust; January 14th 2010 at 06:06 AM.
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  7. #7
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    Just what I thought, but how did you came up with  x_{0}^2 = 20 + r ^2 I'm really sorry for asking too much questions.


    just how did you made x-y=2 to that one
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  8. #8
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    Talking

    Quote Originally Posted by dissidia View Post
    Just what I thought, but how did you came up with  x_{0}^2 = 20 + r ^2 I'm really sorry for asking too much questions.


    just how did you made x-y=2 to that one
    there is a mistake,..I have edited my solution,..see it again,..

    sorry for the mistake
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  9. #9
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    Can you please tell me how did you get (x_{0} - 2)^2 = 8 + r ^2 from x-y=2?
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  10. #10
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    Smile

    Quote Originally Posted by dissidia View Post
    Can you please tell me how did you get (x_{0} - 2)^2 = 8 + r ^2 from x-y=2?
    see the attachment for the picture

    d^2 = (4-2)^2 + 2 ^2

    s = x_{0} - 2

    r^2 + d^2 = s^2
    Attached Files Attached Files
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  11. #11
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    Thank you.
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  12. #12
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    another solution

    Quote Originally Posted by dissidia View Post
    find the eq. Of the circle:
    "the circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

    Can someone help me with this one, i really don't know how to start things up. To be exact, i don't understand the question, sorry.

    Thank you.
    Attached Thumbnails Attached Thumbnails Circle tangent to the line.-eq1.jpg  
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