# Finding dy/dx for a circle

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1. Jul 4, 2017

### MPZ

1. The problem statement, all variables and given/known data
Hello
I have this circle with the equation :

(x-a)^2+(y-b)^2=r^2
I want to find dy/dx for it

2. Relevant equations
(x-a)^2+(y-b)^2=r^2

3. 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!

2. Jul 4, 2017

### 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 ?

3. Jul 4, 2017

### MPZ

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!

4. Jul 4, 2017

### BvU

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 $dy\over dx$ ?

5. Jul 4, 2017

### MPZ

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

6. Jul 4, 2017

### LCKurtz

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

7. Jul 4, 2017

### 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.