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

1. ## 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!

2. ## 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.)

3. ## Re: Finding the tangent to a circle equation given the circle equation

Originally Posted by Rajveer97
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.

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

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

4. ## Re: Finding the tangent to a circle equation given the circle equation

Originally Posted by HallsofIvy
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.

5. ## Re: Finding the tangent to a circle equation given the circle equation

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

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

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

I tried to follow your working but the final answer doesn't seem right from what I saw in the solution?

6. ## Re: Finding the tangent to a circle equation given the circle equation

Originally Posted by Rajveer97
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.

7. ## Re: Finding the tangent to a circle equation given the circle equation

Originally Posted by Plato
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!