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    The line y+4x=k is a tangent of a circle

    a)substitute y=k-4x into the equation of the circle x^2 +y^2=17

    i get x^2 +k^2-8kx+16x^2=17

    b) write down the condition for the equation to have coincident roots

    c)solve the quadratic equation in k to give the 2 values of k and hence the equations of the 2 possible tangents
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    edit: hmmm i think i got it wrong
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    (Original post by Cortez)
    hmmm i think i got it wrong
    join the club
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    No he is right, its kx
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    Your answer to part a) is right i think. I worked it out and got -8x instead of -8kx but ive done it again and got -8kx.

    Im not sure about the others :/
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    I think it should be 16x^2...not 16x
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    so it should be 17x2 +k2 -8kx
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    (Original post by fisfos815)
    so it should be 17x2 +k2 -8kx
    yes sorry for that error,
    can you do part b and c
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    Well for this shouldnt you find the derivative of the circle and find when the gradient = 4

    x^2 + y^2 -17 = 0
    2x + 2y(dy/dx) =0
    dy/dx = -2x/2y

    So when dy/dx = 4
    -2x = 8y
    y= -¼x

    -¼x= k-4x

    x= 4/15 k

    x^2 +k^2-8kx+16x²=17
    (16/225)k² + k² - 32/15 k² +256/225k²-17=0

    16k²+225k²-480k² +256k²-3825=0
    17k²-3825=0

    k²= 3825/17= 225
    k= √225 = ±15
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    I have a stupid question....by coincident do you mean equal roots or real roots?
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    If it means equal roots and I think it does.....
    b2-4ac=0
    (-8k)2-4(17)(k2-17)=0
    64k2-68k2+1156=0
    -4k2=-1156
    k2=289
    k= ±17
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    so the 2 equations are...
    y=17-4x

    y= -4x-17

    .........I think plz correct me if I'm wrong
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    (Original post by fisfos815)
    so the 2 equations are...
    y=17-4x

    y= -4x-17

    .........I think plz correct me if I'm wrong
    yep.
 
 
 
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