# Differential equation of a circle

1. Oct 22, 2012

### iVenky

Consider a circle of radius 'a' and centre (h,b)

then the equation of the circle is given by (x-h)2 + (y-b)2 = a2

I expressed this in terms of differential equations which is -

a= {[1+(dy/dx)2]3/2}/{d2y/dx2}
According to my book - this equation indicates that 'a' is a constant. How can you infer from this equation that 'a' is a constant? (Note: I know that 'a' is a constant but I can't understand how you can infer that from this differential equation).

Last edited: Oct 22, 2012
2. Oct 22, 2012

### iVenky

One more doubt- I tried using Latex for representing that diff equation as one divided by the other but it seems to be working only for numbers and not for differentiation

3. Oct 22, 2012

### haruspex

I think that should be +, not -.
The equation cannot imply that the radius, a, is a constant. In general, it could be a function of x and or y.

4. Oct 22, 2012

### iVenky

Ya it is +. You are sure that this equation can't imply that it is a constant only?

5. Oct 23, 2012

### Staff: Mentor

The right hand side of this equation is the relationship for the local radius of curvature of an arbitrary curve within the x-y plane. Do you know the definition of the curvature or its inverse, the radius of curvature?