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Math Help - Equation of a tangent in a circle

  1. #1
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    Equation of a tangent in a circle

    I am completely horrible at math so I'm not even sure if this is in the right category.
    It's in the algebra section in my math paper anyway.
    Here's the question.

    Given the lne y=kx+3 is a tangent to the circle y^2+x^2=7 determine the possible values of K.
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  2. #2
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    hi, im new but i'll try to help with my knowledge even though its not very good so correct me if i am wrong.

    if the line y=kx+3 is a tangent to the circle, then it has one root.
    so solving y=kx+3 simultaneously with y^2 + x^2 = 7

    we get
    (kx+3)^2 + x^2 = 7
    which simplifies to
    (x^2)(K^2 +1) + 6kx + 2 = 0
    a=k^2 +1 b=6k c=2
    so if we take the discriminant.

    using b^2 - 4ac
    (6k)^2 - 8(K^2 +1)

    28K^2 - 8 = 0

    K^2 = 8/28

    K =
    +(8/28)^(1/2)
    -(8/28)^(1/2)
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  3. #3
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    Uhmm naw I think that's wrong.
    The anser suggests that it's -sqrt2/7

    From (1+kx^2)x^2+6kx+2=o
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  4. #4
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    Quote Originally Posted by Mooki View Post
    Uhmm naw I think that's wrong.
    The anser suggests that it's -sqrt2/7

    From (1+kx^2)x^2+6kx+2=o
    Well, the solution you've given is correct. At the level suggested by this question you're well and truly expected to recognise that \sqrt{\frac{8}{28}} is equivalent to \sqrt{\frac{2}{7}}.

    The only thing that has not been given to you (and it's a very small thing) is which of the two values to keep or in fact whether both values are kept. That's something left for you to determine.
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  5. #5
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    Ok so I'm slightly grasping this question, however I'm stuck as to how (kx+3)^2+x^2=7 simplifies to
    (x^2)(k^2+1)+6kx+2=0

    I mean, it doesn't make sense to me where the 1 came from in (k^2+1). Sorry my math is real horrible, must be embarassing myself here.
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  6. #6
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    hi, its simply expanding the terms and recollecting and then factorising. heres the process (i learnt how to use latex finally!)

    equation we got from solving intersections  (kx+3)^2 +x^2 = 7

    we expand the brackets in  (kx+3)^2 so,
     (kx+3)^2 = [(kx)^2 +6kx + 9]

    now we have  [(kx)^2 +6kx + 9] + x^2 = 7

    now move things around with highest degree term first so you can factorise (if you can)  [(kx)^2 + x^2 +6kx + 9 - 7 = 0

    now notice you can factorise the x^2 out?

     [x^2 (k+1) +6kx + 2 = 0

    its not very hard, you just need to understand a bit more about algerbra - it comes from experience for most people i.e practice
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