1. ## [SOLVED] Solution required...

Please form the differential equation for the circle with centre (h,k) and radius r :-

(x-h)^2 + (y-k)^2 = r^2

answer needs to be in the terms of r ,y and its derivatives. I solved it but it gets too complicated at the end. The equations were getting longer and longer and the end of that half-hour exercise the answer came wrong.

2. Originally Posted by Altair
Please form the differential equation for the circle with centre (h,k) and radius r :-

(x-h)^2 + (y-k)^2 = r^2

answer needs to be in the terms of r ,y and its derivatives. I solved it but it gets too complicated at the end. The equations were getting longer and longer and the end of that half-hour exercise the answer came wrong.
Use implicit differentiation with respect to x on both sides of the circle equation:

$2(x - h) + 2(y - k) \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = - \frac{(x - h)}{(y - k)}$.

Now substitute $(x - h) = \pm \sqrt{r^2 - (y - k)^2}$ .....

3. Thanks. Now the question is that you must have applied implicit differentiation intuitively.

1. How will I know when to and when not to apply it?
2.And by opening and then differentiating the function also comes under implicit differentiation, right?
3.Why does that tend to become more complicated?
4.And can I use implicit differentiation in all such expressions?

I'll be able to understand it a lot better if you can please answer all the above questions.

4. Originally Posted by Altair
1. How will I know when to and when not to apply it?
It's usually used when you can't write y as a function of x, like in this one.

Originally Posted by Altair
2.And by opening and then differentiating the function also comes under implicit differentiation, right?
You can apply it before or after expanding the function.

Originally Posted by Altair
3.Why does that tend to become more complicated?
I think it looks quite simple. What do you mean by complicated?

Originally Posted by Altair
4.And can I use implicit differentiation in all such expressions?
Why not? As long as you look for y' and have an equation, you can use it.

5. Originally Posted by wingless
Why not? As long as you look for y' and have an equation, you can use it.
And for $y''$ as well, right?