Finding the tangent to a circle equation given the circle equation

Jul 2016
10
0
UK
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! :)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.)
 
Apr 2014
7
2
United States
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:
c1.JPG

Substitute x->(x+1) and y->(y+1)
Graph with same having eliminated nonlinear terms
c2.JPG

Move tangent 4y-2x=0, back and plot with original
Substitute x->(x-1) and y->(y-1)
c3.JPG

Answer: 4y-2x-2=0
 
Last edited:
Jul 2016
10
0
UK
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.
 
Jul 2016
10
0
UK
This does not require calculus and is posted
in the Geometry Forum.


Your Problem:
View attachment 35958

Substitute x->(x+1) and y->(y+1)
Graph with same having eliminated nonlinear terms
View attachment 35959

Move tangent 4y-2x=0, back and plot with original
Substitute x->(x-1) and y->(y-1)
View attachment 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?
 

Plato

MHF Helper
Aug 2006
22,489
8,652
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:
Jul 2016
10
0
UK
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!