Equations of Tangents for x^2+(y-4)^2=4 Passing Through the Origin

[Oerg]

Homework Statement

Find the equations of the tangents of the equation $$x^2+(y-4)^2=4$$ that pass through the origin.

The Attempt at a Solution

Ok, I don't know if I am overcomplicating this (takes a deep breath):

The equation of tangent that pass through the origin has the form
$$y=mx$$
And the derivative of the curve is given by
$$\frac{dx}{dy}=\frac{1}{2}(4-(y-4)^2)^\frac{-1}{2}(2y+8)$$
$$\frac{dy}{dx}=\frac{2x}{2y+8}$$
Then equate m=dy/dx and y=mx into the equation of the curve

Here is where it gets really confusing and where I start to doubt my workings.

 

[Defennder]

Well, first of all though not strictly necessary, do note that the graph of the equation is actually a circle of radius 2 centred at (0,4). Once you get that, you should be able to visualise intuitively that because the circle is centred on y-axis, there should be only 2 tangent lines to the circle that passes through the origin.

I don't think your working for dy/dx is correct. Differentiate the equation implicitly, don't bother expressing everything in terms of one variable x or y only. Then you'll be able to do the question easily.

 

[Oerg]

ok so i get

$$2x+2\frac{dy}{dx}-8\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{1}{3}x$$

and then the curve becomes

$$x^2+\frac{1}{9}x^2-\frac{8}{3}x+12=0$$

so i solve for x which gives me the intercepts and i can solve for m is this correct?

 

[blochwave]

Actually impressively I believe his equation for dy/dx is almost right, or I screwed it up(possible!)

if you do it implicity, you get 2x+2(y-4)dy/dx=0? so dy/dx=-2x/(2y-8)

Unless I interpret this wrong though, that means the slope of the tangent lines at the origin is 0, which is obviously not the case, so I think I suck
Edit: You slipped your post in before mine

I believe the error you made is that the derivative of y(x)^2 is 2y*dy/dx, not just 2*dy/dx. With that in mind I think you get the same thing I did

 

[Defennder]

Um no. Differentiating implicitly yields
$$2x + 2(y-4) \frac{dy}{dx} = 0$$
Treat y as a function of x, and remember you are differentiating implicitly with respect to x.

EDIT: Latex doesn't seem to be displaying the equation correctly. Blochwave's answer is correct.

 

[Oerg]

ugh scrap that careless mistakes

so differentiating implicity yields

$$2x+2y\frac{dy}{dx}-8\frac{dy}{dx}=0$$

hey i think i get back the same derivative as my OP

 

[Defennder]

Where did you get $$2y\frac{dy}{dx} - 8\frac{dy}{dx}$$ from?

Implicit differentiation should give $$2x + 2(y-4) \frac{dy}{dx} = 0$$ instead. You're using the chain rule here.

 

[blochwave]

He's right, he just skipped the chain rule and multiplied out that binomial squared

if you factor out the dy/dx you get yours, and mine, answer

 

[Defennder]

Oh yeah. Sheesh. Stupid mistake.

 

[Oerg]

ok so i substitute the derivative and I have this ugly monster

$$x^2+(\frac{2x^2}{2y+8})^2-8(\frac{2x^2}{2y+8})+12=0$$

I have no idea how to simplfiy this, which was why I thought my answer was wrong

 

[Defennder]

How did you get that equation? You substituted the derivative dy/dx into where?