differential equation of circles

**AwaisAmin** · February 20th 2011, 12:41 AM

i have been given the assignment to form the differential equation of circle...
if anyone can please help me...i'll be very thankfu to him...

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..

i dont even know whatst the equation of circle that passes through origin...please help...

**HallsofIvy** · February 20th 2011, 03:08 AM

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.

**AwaisAmin** · February 20th 2011, 04:08 AM

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

**HallsofIvy** · February 20th 2011, 07:48 AM

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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