# Thread: differential equation of circles

1. ## differential equation of circles

i have been given the assignment to form the differential equation of circle...

first..i have to find out the differential equation of a circle with radius r...whose equaton is (x-a)^2 + (y-b)^2 = r^2..
and secondly..i have to find out the differential equation of a circle that passes through origin..

2. The equation of an arbitrary circle includes three independent constants just as you give- and that means you will need a third degree equation to get those arbitrary constant as "constants of integration".

Just differentiating $(x- a)^2+ (y- b)^2= r^2$, with respect to x, gives $2(x- a)+ 2(y- b)\frac{dy}{dx}= 0$ and that has already eliminated a. Differentiating again, $2+ 2\frac{dy}{dx}+ 2(y- b)\frac{d^2y}{dx^2}= 0$ and we have eliminated a. How can you eliminate b?

You can get the equation of a circle passing through the origin by putting x=0, y=0 into the general equation:
$(0- a)^2+ (0- b)^2= a^2+ b^2= r^2$ so that the equation of a circle passing through the origin is
$(x- a)^2+ (y- b)^2= a^2+ b^2$ which has two arbitrary constants.

3. yes..that is my question...how to eliminate b?

4. I had a slight error in my first response. The derivative of $(y- b)\frac{dy}{dx}$, by the product rule, is
$\frac{d(y- b)}{dx}\frac{dy}{dx}+ (y- b)\frac{d^2y}{dx^2}$ so the second differentiation gives
$2+ 2\left(\frac{dy}{dx}\right)^2+ 2(y-b)\frac{d^2y}{dx^2}= 0$. Of course, we can divide through by 2 to get
$1+ \left(\frac{dy}{dx}\right)^2+ (y- b)\frac{d^2y}{dx^2}= 0$.

And, we can write that as
$bfrac{d^2y}{dx^2}= 1+ \left(\frac{dy}{dx}\right)^2+ y\frac{d^2y}{dx^2}$
so that
$b= \frac{ 1+ \left(\frac{dy}{dx}\right)^2+ y\frac{d^2y}{dx^2}}{\frac{d^2y}{dx^2}}$

Differentiating for a third time,
$2\frac{dy}{dx}\frac{d^2y}{dx^2}+ \frac{dy}{dx}\frac{d^2y}{dx^2}+ (y- b)\frac{d^3y}{dx^3}= 0$

Now you can replace the "b" in that from the previous equation and simplify.

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# find differential equation of (x-a)² (y-b)² = r⅔

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