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    Absolutely no idea how to do this one...
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    (Original post by Retsek)
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    Absolutely no idea how to do this one...
    Are you familiar with the factorised form of the equation of a circle?

    I.e.  (x-h)^2 + (y-k)^2 = r^2
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    (Original post by Retsek)
    Absolutely no idea how to do this one...
    3(x^2+y^2-4x-3y)+1=0

    Now complete the squares on x and y inside the bracket.
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    (Original post by Retsek)
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    Absolutely no idea how to do this one...
    Also this one, is it just trial and error?
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    (Original post by RDKGames)
    3(x^2+y^2-4x-3y)+1=0

    Now complete the squares on x and y inside the bracket.
    Thanks a lot! I got that far before getting myself confused
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    (Original post by Retsek)
    Also this one, is it just trial and error?
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    Start visualising and drawing sketches.

    if one of those lines are a tangent to the circle, then you know that one of those lines touch the circle at a single point. Bang all of them in to the circle equation and find out which touch the circle at a single point
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    (Original post by Retsek)
    Thanks a lot! I got that far before getting myself confused
    3(x^2+y^2-4x-3y)+1=0
    3((x^2-4x)+(y^2-3y))+1=0

    You know about completing the square right? As RDK said, just do that on the two inner brackets, and you'll get the answer (after some rearranging).
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    (Original post by Retsek)
    Also this one, is it just trial and error?
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    (Original post by Naruke)
    Start visualising and drawing sketches.

    if one of those lines are a tangent to the circle, then you know that one of those lines touch the circle at a single point. Bang all of them in to the circle equation and find out which touch the circle at a single point
    While this is rigorous and bound to give the correct answer, a quicker approach is to notice that the radius is 2 and centre at the origin - and so you're looking for a line which has a minimum distance from the origin as 2.

    For example; the fourth equation fails this because the line crosses through (0,\sqrt{2}) and (\sqrt{2},0) and since \sqrt{2}<2 you have the line going through 2 axis points which are inside the circle, with minimum distance being less than 2 as a result.
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    (Original post by Darth_Narwhale)
    3(x^2+y^2-4x-3y)+1=0
    3((x^2-4x)+(y^2-3y))+1=0

    You know about completing the square right? As RDK said, just do that on the two inner brackets, and you'll get the answer (after some rearranging).
    Yep I got it now, thanks man
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    (Original post by RDKGames)
    While this is rigorous and bound to give the correct answer, a quicker approach is to notice that the radius is 2 and centre at the origin - and so you're looking for a line which has a minimum distance from the origin as 2.

    For example; the fourth equation fails this because the line crosses through (0,\sqrt{2}) and (\sqrt{2},0) and since \sqrt{2}<2 you have the line going through 2 axis points which are inside the circle, with minimum distance being less than 2 as a result.
    Oh I see now, cheers
 
 
 
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