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Math Help - circles,tangents problem...

  1. #1
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    circles,tangents problem...

    hello guys a new member here..
    i need help with this math problem that i got recently... its in the chapter conic sections.
    the question is:
    Find the equation of the tangent to the circle
    if the equation of the circle is give by x^2+y^2-4x+6y-12=0 and is parallel to the line having the equation given by x+y-8=0

    any help would be appreciated ..
    cheers.
    EDIT: last question was wrong and i updated it with a valid question.
    thx for the response.
    Last edited by GalXiOn; December 13th 2011 at 08:19 AM.
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  2. #2
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    Re: circles,tangents problem...

    Quote Originally Posted by GalXiOn View Post
    Find the equation of the tangent to the circle
    Given: eq. of circle - x^2-4x-6y-12=0 and tangent is parallel to the line given by the equation 4x-3y=1.
    There is something wrong with the question.
    x^2-4x-6y-12=0 is not a circle.
    Please check the wording of that question.
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  3. #3
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    Re: circles,tangents problem...

    Quote Originally Posted by Plato View Post
    There is something wrong with the question.
    x^2-4x-6y-12=0 is not a circle.
    Please check the wording of that question.
    umm in my book it says if the equation if of the form x^2+y^2+2gx+2fy+c=0 then it is a General form i.e it represents the circle.
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  4. #4
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    Re: circles,tangents problem...

    Quote Originally Posted by GalXiOn View Post
    umm in my book it says if the equation if of the form x^2+y^2+2gx+2fy+c=0 then it is a General form i.e it represents the circle.
    But what you posted is not a circle there is no y^2~.
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  5. #5
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    Re: circles,tangents problem...

    Quote Originally Posted by Plato View Post
    But what you posted is not a circle there is no y^2~.
    EDIT: updated.... (question given was wrong Sorry.. updated now...)
    Last edited by GalXiOn; December 13th 2011 at 08:15 AM.
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  6. #6
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    Re: circles,tangents problem...

    well if not that,what about this one? :
    x^2+y^2-4x+6y-12=0 and parallel to line having the equation x+y-8=0.. will this work?.. im just listing out a few sums that i got recently.. ive never seen these sorta sums before.. and are totally new.. im guessing its gotta do something with comparing the radius using 2 different formulas to get a variable.. something like equating (|ax1+by1+c|)/ sqrt(A^2+b^2) which is the distance from a line and a point
    and this formula: sqrt(g^2+f^2-c) which is the radius of the circle if g,f,c is known to us(which can be found using the given equation) g is -2 and f is -3..(im not sure..)
    now i dont know how to proceed or i dont know if it is even the right way to approach.. ..
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  7. #7
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    Re: circles,tangents problem...

    Quote Originally Posted by GalXiOn View Post
    well if not that,what about this one? :
    x^2+y^2-4x+6y-12=0 and parallel to line having the equation x+y-8=0..
    By completing squares that circle is (x-2)^2+(y+3)^2=25.
    The slope of any non-vertical tangent is m=\frac{-(a-2)}{(b+3)} where (a-2)^2+(b+3)^2=25 and b\ne -3.

    The slope of the line x+y-8=0 is -1~.

    So find points (a,b) where slope is -1.
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  8. #8
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    Re: circles,tangents problem...

    Quote Originally Posted by Plato View Post
    By completing squares that circle is (x-2)^2+(y+3)^2=25.
    The slope of any non-vertical tangent is m=\frac{-(a-2)}{(b+3)} where (a-2)^2+(b+3)^2=25 and b\ne -3.

    The slope of the line x+y-8=0 is -1~.

    So find points (a,b) where slope is -1.
    hmm thx i'll try it out..
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