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Math Help - I'm new to this forum...can someone help with a function problem?

  1. #1
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    I'm new to this forum...can someone help with a function problem?

    The problem is:

    Find an equation for the family of line tangent to the circle with the centre at the origin and radius 3.

    I've been looking at this problem all day and can't for the life of me crack it! I would really appreciate some help or I won't sleep tonight!!
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by lindax View Post
    The problem is:

    Find an equation for the family of line tangent to the circle with the centre at the origin and radius 3.

    I've been looking at this problem all day and can't for the life of me crack it! I would really appreciate some help or I won't sleep tonight!!
    The equation of the circle is:

    x^2+y^2=9

    The equation of the line is y=mx+c

    Now, substitute the second equation into the first, and make it so there is only one solution for x



    My Solution:

    x^2+(mx+c)^2=9

    x^2+m^2x^2+2mxc+c^2=9

    (m^2+1)x^2+(2mc)x+(c^2-9)=0

    \Delta = (2mc)^2-4(m^2+1)(c^2-9)

    For there to be one solution, \Delta =0

    4m^2c^2-(4m^2c^2+4c^2-36m^2-36)=0

    36m^2-4c^2+36=0

    36m^2=4c^2-36

    m^2 = \frac{1}{9}(c^2-9)

    m=\pm \sqrt{\frac{1}{9}(c^2-9})

    m=\pm \frac{\sqrt{c^2-9}}{3}

    Family is:

    \lbrace{(x,y):y = \pm \frac{\sqrt{c^2-9}}{3}x + c, c^2-9 \geq 0\rbrace}

    If you want you could also describe this in terms of m
    Last edited by DivideBy0; December 14th 2007 at 08:45 PM.
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  3. #3
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    Thanks Divideby0

    I really like your answer and understand it but it's not the one in the back of the textbook which is:

    y = 9-x0x/9- x0

    Does this make any sense to you? I am stumped....

    cheers
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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by lindax View Post
    Thanks Divideby0

    I really like your answer and understand it but it's not the one in the back of the textbook which is:

    y = 9-x0x/9- x0

    Does this make any sense to you? I am stumped....

    cheers
    Sorry, I can't quite understand the equation you have there. What's with the 0's?
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  5. #5
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    hmm they don't seem to look right do they....

    the 0's are supposed to be subscripts - sorry!

    (By they way, how do you make equations look acceptable on here?)

    thanks for persevering with my question, Divideby0 - I really do appreciate your help.

    cheers

    L
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  6. #6
    Senior Member DivideBy0's Avatar
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    You can use the LaTeX ('Lay-Tech') typesetting. There is a thread on how to use it here:

    http://www.mathhelpforum.com/math-he...-tutorial.html

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  7. #7
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    Hello, lindax!

    I don't trust your textbook.
    No one writes the equation of a line like that!


    Find an equation for the family of line tangent to the circle
    with the centre at the origin and radius 3.
    The equation of the circle is: . x^2+y^2\:=\:9

    Differentiate implicitly: . 2x + 2yy' \:=\:0\quad\Rightarrow\quad y' \:=\:-\frac{x}{y}

    The line through (x_o,\,y_o) with slope -\frac{x_o}{y_o} .has the equation:

    . . . y - y_o \:=\:-\frac{x_o}{y_o}(x - x_o)\quad\Rightarrow\quad y \:=\:\frac{-x_ox+x_o^2+y_o^2}{y_o}


    Since: . x_o^2+y_o^2 \:=\:9\:\text{ and }\:y_o \:=\:\pm\sqrt{9-x_o^2}

    . . the equation becomes: . y \:=\:\frac{-x_ox + 9}{\pm\sqrt{9-x_o^2}} \quad\Rightarrow\quad y \;=\;\pm\frac{9-x_ox}{\sqrt{9-x_o^2}}

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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, lindax!

    I don't trust your textbook.
    No one writes the equation of a line like that!


    The equation of the circle is: . x^2+y^2\:=\:9

    Differentiate implicitly: . 2x + 2yy' \:=\:0\quad\Rightarrow\quad y' \:=\:-\frac{x}{y}

    The line through (x_o,\,y_o) with slope -\frac{x_o}{y_o} .has the equation:

    . . . y - y_o \:=\:-\frac{x_o}{y_o}(x - x_o)\quad\Rightarrow\quad y \:=\:\frac{-x_ox+x_o^2+y_o^2}{y_o}


    Since: . x_o^2+y_o^2 \:=\:9\:\text{ and }\:y_o \:=\:\pm\sqrt{9-x_o^2}

    . . the equation becomes: . y \:=\:\frac{-x_ox + 9}{\pm\sqrt{9-x_o^2}} \quad\Rightarrow\quad y \;=\;\pm\frac{9-x_ox}{\sqrt{9-x_o^2}}

    well, that kinda looks like the answer in the text (except for the denominator) but i don't think we're allowed to use calculus here (we're in high school/pre-calculus math). i'd do it pretty much the same way DivideBy0 did it. can you see another way, Soroban?
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  9. #9
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    Hi Soroban

    Thanks so much...I can see now why I could not work out how to get that answer - my calculus is very rusty after a 30 year break! (I'm working on improving it by making my way through this text - Anton, Bivens & Davis -Calculus 7ed.)

    I think that given the context of the question, Divideby0's answer was more appropriate for this level.

    Thank you both for your help.

    Just another question while you are there, if I use LaTex notation for my equations on this forum, should I see it when I do a preview of the post or does it show up afterwards when the message has been submitted?

    L
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lindax View Post
    Just another question while you are there, if I use LaTex notation for my equations on this forum, should I see it when I do a preview of the post or does it show up afterwards when the message has been submitted?

    L
    Yes, LaTeX will show up when you preview your post
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