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    There is a very similar question in an AEA paper I believe


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    (Original post by BuryMathsTutor)
    If, for a given value of m, the quadratic has equal roots then this value produces a line which is a tangent to the circle. You need to solve discriminant = 0.

    You have a small error in your equation by the way. The -4 should be +4.

    You should get a quadratic in m: 44m^2+96m+16=0
    or
    11m^2+24m+4=0
    Thanks for your reply.

    I get:

    100m^4 + 224m^3 + 116m^2 + 96m + 16 = 0

    how did you end up with no m^4 or m^3 terms given the above x based quadratic?

    I must be missing something pretty basic!
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    (Original post by Sm0key)
    Thanks for your reply.

    I get:

    100m^4 + 224m^3 + 116m^2 + 96m + 16 = 0

    how did you end up with no m^4 or m^3 terms given the above x based quadratic?

    I must be missing something pretty basic!
    \displaystyle (1 + m^2)x^2 + (8m^2 + 6m)x + (16m^2 + 24m + 4) has discriminant

    \displaystyle \Delta = (8m^2 + 6m)^2 - 4(1+m^2)(16m^2 + 24m + 4)

    \displastyle = 64m^4 + 96m^3 + 36m^2 - 4(16m^4 + 24m^3 + 20m^2 + 24m + 4)

    \displaystyle = -4(11m^2 + 24m + 4) = -4(m+2)(11m+2)
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    I got y= (-4x+5)/3 for one of them...
    You get the tangent because you use P(x,y) as the point of tangency and get the gradient i.t.o this x,y pair.
    I just solved a pair of sim eqns... is this right?
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    wait a minute i made a slight algebra slip and got it wrong
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    is this C2?
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    (Original post by Patrick2810)
    is this C2?
    I hope not, I don't believe we have implicit differentiation in C2 like this.
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    (Original post by Louisb19)
    I hope not, I don't believe we have implicit differentiation in C2 like this.
    correct me if im wrong but you dont need implicit differentiation if you let the point of tangency be P(x,y) say, and since you have the coords of the midpoint (0,0) you have the gradient of the tangent using -1/m
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    (Original post by MathMeister)
    correct me if im wrong but you dont need implicit differentiation if you let the point of tangency be P(x,y) say, and since you have the coords of the midpoint (0,0) you have the gradient of the tangent using -1/m
    I tried to solve it like that but I ended up with very weird equations which would take forever to solve
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    (Original post by Zacken)
    \displaystyle (1 + m^2)x^2 + (8m^2 + 6m)x + (16m^2 + 24m + 4) has discriminant

    \displaystyle \Delta = (8m^2 + 6m)^2 - 4(1+m^2)(16m^2 + 24m + 4)

    \displastyle = 64m^4 + 96m^3 + 36m^2 - 4(16m^4 + 24m^3 + 20m^2 + 24m + 4)

    \displaystyle = -4(11m^2 + 24m + 4) = -4(m+2)(11m+2)
    Thanks for the working. To confirm my answer with BuryMathsTutor too, my final solutions are:

    y + 2x + 5 = 0
    2x + 11y + 5 = 0

    which doesn't match the post 8 solutions.....
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    if its too hard to solve x^2 +y^2-5=0 and x^2+y^2-3y+4x=0
    and then use the point (-4,3) then use the discriminant method then..
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    (Original post by Sm0key)
    Thanks for the working. To confirm my answer with BuryMathsTutor too, my final solutions are:

    y + 2x + 5 = 0
    2x + 11y + 5 = 0

    which doesn't match the post 8 solutions.....
    The first of these is correct. The other has the correct gradient but does not pass through the point (-4,3). Start with 2x+11y+d=0. Then 2(-4)+11(3)+d=0 to get d.
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    My first instinct was the discriminant method, but there might be a nice geometric method... I'll have a ponder.

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    (Original post by BuryMathsTutor)
    The first of these is correct. The other has the correct gradient but does not pass through the point (-4,3). Start with 2x+11y+d=0. Then 2(-4)+11(3)+d=0 to get d.
    Whoops yes I made an arithmetic slip. So my final solutions are in fact:

    y + 2x + 5 = 0

    2x + 11y - 25 = 0

    Correct I trust?
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    (Original post by Sm0key)
    Whoops yes I made an arithmetic slip. So my final solutions are in fact:

    y + 2x + 5 = 0

    2x + 11y - 25 = 0

    Correct I trust?

    :yes:
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    (Original post by BuryMathsTutor)
    :yes:
    Great, thanks a lot for your help.
 
 
 
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