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Thread: Finding the tangent to a circle equation given the circle equation

  1. #1
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    Exclamation Finding the tangent to a circle equation given the circle equation

    Hi guys, first time poster here!

    There' this equation I don't quite understand. I'm given the following question:

    The tangent line to the curve (x^2) (y^2) − (2x^2) + (y^2) = 0 at the point (1, 1) is=?

    Not quite sure how to deal with the circle arranged that way? I really need to understand this and fast

    Thank you for any help!
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  2. #2
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    Re: Finding the tangent to a circle equation given the circle equation

    Your real problem is that this curve is NOT a circle! The tangent line to any smooth curve, at a given point, is a line that goes through that point and has slope equal to the derivative at that point. The derivative of y, given by $\displaystyle x^2y^2- 2x^2+ y^2= 0$, is given by $\displaystyle 2xy^2- 2x^2y \frac{dy}{dx}- 4x+2y\frac{dy}{dx}= 0$. At the point (1, 1), setting x= y= 1, $\displaystyle 2- 2\frac{dy}{dx}- 4+ 2\frac{dy}{dx}= 0$ so $\displaystyle 2- 4= 0$. The $\displaystyle \frac{dy}{dx}$ completely cancels out leaving a untrue equation! That means that the derivative does not exist there which, in turn, means the tangent line is vertical. What is the equation of a vertical line that passes through (1, 1)?

    (You seem to be saying that you thought that only circles can have "tangent lines". If so, you have missed a lot of material that is preliminary to problems like this. You need to talk to your teacher about this.)
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    Re: Finding the tangent to a circle equation given the circle equation

    Quote Originally Posted by Rajveer97 View Post
    Hi guys, first time poster here!

    There' this equation I don't quite understand. I'm given the following question:

    The tangent line to the curve (x^2) (y^2) − (2x^2) + (y^2) = 0 at the point (1, 1) is=?

    Not quite sure how to deal with the circle arranged that way? I really need to understand this and fast

    Thank you for any help!
    This does not require calculus and is posted
    in the Geometry Forum.


    Your Problem:
    Finding the tangent to a circle equation given the circle equation-c1.jpg

    Substitute x->(x+1) and y->(y+1)
    Graph with same having eliminated nonlinear terms
    Finding the tangent to a circle equation given the circle equation-c2.jpg

    Move tangent 4y-2x=0, back and plot with original
    Substitute x->(x-1) and y->(y-1)
    Finding the tangent to a circle equation given the circle equation-c3.jpg

    Answer: 4y-2x-2=0
    Last edited by RLBrown; Jul 5th 2016 at 12:18 PM.
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    Re: Finding the tangent to a circle equation given the circle equation

    Quote Originally Posted by HallsofIvy View Post
    Your real problem is that this curve is NOT a circle! The tangent line to any smooth curve, at a given point, is a line that goes through that point and has slope equal to the derivative at that point. The derivative of y, given by $\displaystyle x^2y^2- 2x^2+ y^2= 0$, is given by $\displaystyle 2xy^2- 2x^2y \frac{dy}{dx}- 4x+2y\frac{dy}{dx}= 0$. At the point (1, 1), setting x= y= 1, $\displaystyle 2- 2\frac{dy}{dx}- 4+ 2\frac{dy}{dx}= 0$ so $\displaystyle 2- 4= 0$. The $\displaystyle \frac{dy}{dx}$ completely cancels out leaving a untrue equation! That means that the derivative does not exist there which, in turn, means the tangent line is vertical. What is the equation of a vertical line that passes through (1, 1)?

    (You seem to be saying that you thought that only circles can have "tangent lines". If so, you have missed a lot of material that is preliminary to problems like this. You need to talk to your teacher about this.)
    I spent some time trying to make sense of your explanation and turns out I had pretty much missed the entire topic of Implicit functions, saying circle instead of curve was a mistake, I should have spotted the difference in the type of equations that describe a circle or curve. If you don't mind could you explain to me how you got that equation? Because when I tried to do it again myself after learning implicit function I got (2x^2)y + (2xy^2) - 4x +2y = 0. Also sorry if this is a silly question but how does one know the tangent line is vertical if the derivative doesn't exist? This is the topic that's really bugging me right now, my knowledge on it is quite weak since I missed. Any help would be great. Thanks again.
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  5. #5
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    Re: Finding the tangent to a circle equation given the circle equation

    Quote Originally Posted by RLBrown View Post
    This does not require calculus and is posted
    in the Geometry Forum.


    Your Problem:
    Click image for larger version. 

Name:	c1.JPG 
Views:	3 
Size:	28.2 KB 
ID:	35958

    Substitute x->(x+1) and y->(y+1)
    Graph with same having eliminated nonlinear terms
    Click image for larger version. 

Name:	c2.JPG 
Views:	5 
Size:	49.7 KB 
ID:	35959

    Move tangent 4y-2x=0, back and plot with original
    Substitute x->(x-1) and y->(y-1)
    Click image for larger version. 

Name:	c3.JPG 
Views:	4 
Size:	33.8 KB 
ID:	35960

    Answer: 4y-2x-2=0
    Thanks for your reply
    I tried to follow your working but the final answer doesn't seem right from what I saw in the solution?
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  6. #6
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    Re: Finding the tangent to a circle equation given the circle equation

    Quote Originally Posted by Rajveer97 View Post
    The tangent line to the curve (x^2) (y^2) − (2x^2) + (y^2) = 0 at the point (1, 1) is=?
    There is a simple sign error in Prof Ivy's calculation.

    I get $2xy^2+2x^2yy'-4x+2yy'=0$ as the implicit derivative. So at the point $(1,1)$ we have
    $\displaystyle \begin{align*}2+2y'-4+2y'&=0 \\4y'&=2\\y'&=\frac{1}{2} \end{align*}$

    Now what is the the answer? See here.
    Last edited by Plato; Jul 6th 2016 at 02:11 PM.
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  7. #7
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    Re: Finding the tangent to a circle equation given the circle equation

    Quote Originally Posted by Plato View Post
    There is a simple sign error in Prof Ivy's calculation.

    I get $2xy^2+2x^2yy'-4x+2yy'=0$ as the implicit derivative. So at the point $(1,1)$ we have
    $\displaystyle \begin{align*}2+2y'-4+2y'&=0 \\4y'&=2\\y'&=\frac{1}{2} \end{align*}$

    Now what is the the answer? See here.
    Yep I noticed that just a while and managed to get the right equation: 2y-x-1=0 which is also the right answer. Thanks for the help!
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