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    Hi, I'm stuck on this question as I'm just unsure of how to progress, I'll show my working out and hopefully one of you nice people (or a few! ) can help

    "The straight line with equation y=kx is a tangent to the circle x^2 + y^2 - 4x - 4y + 7 = 0. Find the possible values of k, giving your answers in the form a+b7^1/2."

    So I substituted y=kx into the circle formula and got:

    x^2 + (kx)^2 - 4x - 4(kx) + 7 =0

    So: x^2 +k^2x^2 - 4x - 4kx + 7 = 0

    x^2 (1+k^2) - 4x (1+k^2) + 7 = 0

    And then I'm not sure what to do. I was going to use the discriminant (as it's a tangent b^2-4ac will = 0, but then I'd get a k^2*k^2 getting k^4? )

    This is very similar to another question "The straight line with equation y=-3x+c is a tangent to the circle x^2+y^2-4x-2y-5=0. Find the possible values of c", but hopefully after understanding 1. I shall be able to do 2 as well Thanks!
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    There is no k^4

    You have turned a k into a k^2 in the x part
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    (Original post by TenOfThem)
    There is no k^4

    You have turned a k into a k^2 in the x part
    :O oh yeah! So am I doing it right?
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    yes
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    (Original post by TenOfThem)
    yes
    Erm still, it's wrong. As Then I have:

    (1-k)^2 - 4(1+k^2)(7) = 0

    So 1 + k^2 - 2k - 28 + 28k^2 = 0

    29k^2 - 2k -27 =0

    Quadratic formula...I then get 2+/-6 root87 (over) 58.

    But it's meant to be 4/3 +/- 1/3 root7 ??
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    x^2 + k^2x^2 - 4x - 4kx + 7 = (k^2+1)x^2 - 4(k+1)x + 7



b^2-4ac = 16(k+1)^2 - 28(k^2+1) = 16k^2 + 32k + 16 - 28k^2 - 28 



so 12k^2 - 32k +12 = 0



so 3k^2 - 8k + 3 = 0
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k =  {(8\pm\sqrt{64-36})}/{6}



= [8\pm2\sqrt7]/6

    giving as req
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    (Original post by TenOfThem)
    x^2 + k^2x^2 - 4x - 4kx + 7 = (k^2+1)x^2 - 4(k+1)x + 7



b^2-4ac = 16(k+1)^2 - 28(k^2+1) = 16k^2 + 32k + 16 - 28k^2 - 28 



so 12k^2 - 32k +12 = 0



so 3k^2 - 8k + 3 = 0
    Ah, thanks so I think it's the 4(k+1)x, as I forgot the 4. Thanks! I'll see if I can do the second question now
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    (Original post by TenOfThem)
    

k =  {(8\pm\sqrt{64-36})}/{6}



= [8\pm2\sqrt7]/6

    giving as req
    If that was the answer, then couldn't you just divide the 8 by 6 to get 4/3, but then wouldn't it be 4/3 +/- 2 root 7 but the answer is 4/3 =/- 1/3 root 7 (which makes sense if you divide the whole numerator by 6, but don't you just divide one part of it? i.e. either the 4 or the 2root7?
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    (Original post by MedicalMayhem)
    If that was the answer, then couldn't you just divide the 8 by 6 to get 4/3, but then wouldn't it be 4/3 +/- 2 root 7 but the answer is 4/3 =/- 1/3 root 7 (which makes sense if you divide the whole numerator by 6, but don't you just divide one part of it? i.e. either the 4 or the 2root7?
    no you divide the whole lot

    \frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b
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    (Original post by TenOfThem)
    no you divide the whole lot

    \frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b
    Oh yeah sorry, just remembered that you only do that if it is x * y. not x+y.
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    (Original post by TenOfThem)
    no you divide the whole lot

    \frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b
    Erm, could you help with this question as well?

    "The straight line with equation y=-3x+c is a tangent to the circle x^2+y^2-4x-2y-5=0. Find the possible values of c"

    Well I'm just unsure how you would work out the c. As for co-ordinates where the lines intercept you could substiute one into another, but for this? I thought you might find the possible y and x values and then you get c, but I'm not too sure how to get the points? :/

    Would you get the centre of the circle and the radius. So:

    (x-2)^2 + (y-1)^2 -5 (and do you also do -4 - 1 as well, from expanding the brackets? As I remember (i think) that you don't but never learnt why...). So is it (x-2)^2 + (y-1)^2 -5 -4 -1 (or is it without the -4, -1)?

    And then you get the center as 2,1, but then the radius is 0? :/ (as -4,-1 (from the conpleting the square bit) + 5 = 0?)
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    Ok so completing the square does give

    (x-2)^2 - 4 + (y-1)^2 - 1 - 5 = 0
    giving

    (x-2)^2 + (y-1)^2 = 10

    centre = (2,1) and radius =\sqrt{10}
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    But that is not how you do this problem

    Substitute y = -3x+c into the original equation

    You will have a quadratic in x that will give the values where the line and the circle meet

    Since the line is a tangent how many solutions will this equation have?
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    (Original post by TenOfThem)
    But that is not how you do this problem

    Substitute y = -3x+c into the original equation

    You will have a quadratic in x that will give the values where the line and the circle meet

    Since the line is a tangent how many solutions will this equation have?
    1 (the discriminant bit?).

    But with your previous post, isn't the -5 which is already there shows the (-4,-1) from the completing the square? So would it be radius as root5? :/
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    yes the discriminant

    The -5 that is already there is already there

    In the equation you have three bits [x^2-4x], [y^2-2y], and [-5]

     x^2-4x = (x-2)^2-4

y^2-2y = (y-1)^2-1

-5 = -5
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    (Original post by TenOfThem)
    yes the discriminant

    The -5 that is already there is already there

    In the equation you have three bits [x^2-4x], [y^2-2y], and [-5]

     x^2-4x = (x-2)^2-4

y^2-2y = (y-1)^2-1

-5 = -5
    Ah okay, that's what I originally thought, but my other friend confused me. So for that question, I assume you just find the points where the line and curve cross, then substitute into the line (y=-3x+c) to get c. So seems easy enough, but then how would you go about creating a simultaneous equation with a c? :/

    EDIT, or could you just explain what would be the first step to work it out?
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    y=-3x+c



so x^2 - 4x + (-3x+c)^2 - 2(-3x+c) -5 =0



10x^2 + (2-6c)x + (c^2-2c-5) = 0

    A tangent so you only want one solution
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    (Original post by TenOfThem)
    y=-3x+c



so x^2 - 4x + (-3x+c)^2 - 2(-3x+c) -5 =0



10x^2 + (2-6c)x + (c^2-2c-5) = 0

    A tangent so you only want one solution
    Oh so you actually can insert c But why does it say possible 'values' when you say you only want one solution?
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    One solution for x because there is only one point where the tangent and the circle meet

    This will give 2 values of c

    Spoiler:
    Show

    

(2-6c)^2 - 40(c^2-2c-5) = 0



36c^2 - 24c - 40c^2 + 80c + 204 = 0



c^2 - 14c - 51 = 0



(c-17)(c+3)=0
 
 
 
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