What is Implicit Differentiation for a Circle?

[MPZ]

Homework Statement

Hello
I have this circle with the equation : [/B]
(x-a)^2+(y-b)^2=r^2
I want to find dy/dx for it

2. Homework Equations
(x-a)^2+(y-b)^2=r^2

The Attempt at a Solution

I am looking on the internet and it appears that I should use what is called "Implicit differentiation", can anyone use "Implicit differentiation for this circle please, thanks!

 

[BvU]

Hi,
Keep it simple and set a=b=0 . If you differentiate $x^2+y^2 = r^2$ you get $2xdx+2ydy=0$ so $\displaystyle {{dy\over dx}= -{x\over y}}$ which can be checked easily.

PS is this homework ?

 

[MPZ]

Hi,
Keep it simple and set a=b=0 . If you differentiate $x^2+y^2 = r^2$ you get $2xdx+2ydy=0$ so $\displaystyle {{dy\over dx}= -{x\over y}}$ which can be checked easily.

PS is this homework ?

this is not a homework, i am trying to use a mathematical software to draw images using mathematical equations :) Can you please tell me a way to not "keep it simple" since i need the values of a and b since I can't draw it that small!

 

[BvU]

this is not a homework

Fair enough.

not "keep it simple"

The idea was to make it easier to understand, so that you would be able to complicate things on your own. Same differentiation on $(x-a)^2+(y-b)^2 = r^2$ gives you $2(x-a)dx+2(y-b)dy=0$ which is not very surprising if you see it as a translation of the origin to $(a,b)$.

draw images using mathematical equations

Can you elaborate? Give an example why you need $dy\over dx$ ?

 

[MPZ]

Fair enough.
The idea was to make it easier to understand, so that you would be able to complicate things on your own. Same differentiation on $(x-a)^2+(y-b)^2 = r^2$ gives you $2(x-a)dx+2(y-b)dy=0$ which is not very surprising if you see it as a translation of the origin to $(a,b)$.

Can you elaborate? Give an example why you need $dy\over dx$ ?

are you a detective? LOOL. I need the derivative because from it I can get the slope of the normal at any point since I want to find the equation of multiple normal lines at different points to draw "the hair" of the thing I am drawing. No more questions please with this sort

 

[LCKurtz]

You hardly need calculus and derivatives for that. The normal lines to a circle are radius lines. Straight lines through the center.

 

[LCKurtz]

Here's a sample written in Maple:
> restart; with(plots);
> circle := plot([cos(theta), sin(theta), theta = 0 .. 2*Pi], color = black, axes = none, thickness = 2):
> hair := seq(plot([r*cos((1/3)*Pi+(1/36)*k*Pi), r*sin((1/3)*Pi+(1/36)*k*Pi), r = 1 .. 1.2], thickness = 2), k = 0 .. 12);
> display({hair}, circle);

Maybe that will give you some ideas.