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Thread: coordinate geometry, circle

  1. #1
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    coordinate geometry, circle

    Hi there,
    I have this : The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.

    So am i right doing it like this:

    (x-4)2 + (x+a-1)2 = 16
    x2 - 8x + 16 + x2 + a2 + 1 + 2xa - 2x - 2a - 16 = 0
    2x2 - 10x + a2 + 2xa - 2a +1 = 0
    now using b2-4ac:
    (-10 + 2a)2 - 4 (2) (a2 - 2a + 1) which gives me
    -4 (a2 + 6a - 23)
    and that gives me possible values of 23.65 and and 12.34 when i should be getting -8.66 or 2.66.

    Can anyone tell me what im doing wrong ?
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  2. #2
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    Re: coordinate geometry, circle

    Quote Originally Posted by AbYz View Post
    The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.
    You posted this in pre-collage, meaning that calculus is not to be used. Is that correct?
    Do you know other methods to find points on the circle where the slope is $1~?$
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  3. #3
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    Re: coordinate geometry, circle

    Quote Originally Posted by AbYz View Post
    Hi there,
    I have this : The line y = x + a , where a is a constant is a tangent to the circle (x-4)2 + (y-1)2 = 16. Find two possible values of a.

    So am i right doing it like this:

    (x-4)2 + (x+a-1)2 = 16
    x2 - 8x + 16 + x2 + a2 + 1 + 2xa - 2x - 2a - 16 = 0
    2x2 - 10x + a2 + 2xa - 2a +1 = 0
    now using b2-4ac:
    (-10 + 2a)2 - 4 (2) (a2 - 2a + 1) which gives me
    -4 (a2 + 6a - 23)=0
    You are correct up to here. Seems you have made a mistake using the quad formula. I got -8.66 or 2.66. Try it again from here.

    and that gives me possible values of 23.65 and and 12.34 when i should be getting -8.66 or 2.66.

    Can anyone tell me what im doing wrong ?
    see comment in red
    Thanks from AbYz
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  4. #4
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    Re: coordinate geometry, circle

    Yes i did stupid mistake with formula..

    Quote Originally Posted by Plato View Post
    You posted this in pre-collage, meaning that calculus is not to be used. Is that correct?
    Do you know other methods to find points on the circle where the slope is $1~?$
    Hmm , i could not think of anything there to be honest so after googling i did this ^. Can you share with me this other method ? i can only think of simultaneous equations but im not sure of it as i end up with 3 unknowns ( x , y and a) ?
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  5. #5
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    Re: coordinate geometry, circle

    Quote Originally Posted by AbYz View Post
    Yes i did stupid mistake with formula..
    Hmm , i could not think of anything there to be honest so after googling i did this ^. Can you share with me this other method ? i can only think of simultaneous equations but im not sure of it as i end up with 3 unknowns ( x , y and a) ?
    If $(p,q)$ is a point on the circle $(x-h)^2+(y-k)^2=r^2$, then the line determined by the point and the center has slope $m=\dfrac{q-k}{p-h}$, of course there a special case if $p=h$.
    A tangent to the circle at $(p,q)$ has slope $m_{\bot}=-m^{-1}$, because it is perpendicular to the radial segment.

    Thus in the given problem, we look for points in the circle at which the line to center has slope is $-1$ because the given tangent has slope $1$
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  6. #6
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    Re: coordinate geometry, circle

    not sure im following, i have
    -1 = \frac{q-1}{p-4}
    -p+4=q-1\\  \ p=5-q\\  \ q=5-p\\ \ p = 2.5\\ \ q= 2.5\\

    what should i do with this ?
    Last edited by AbYz; Feb 22nd 2017 at 12:34 PM.
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  7. #7
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    Re: coordinate geometry, circle

    Quote Originally Posted by AbYz View Post
    not sure im following, i have
    -1 = \frac{q-1}{p-4}
    -p+4=q-1\\  \ p=5-q
    $((5-q)-2)^2+(q-1)^2=16^2$ SOLVE for $q$ to for $p$
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