Differentiating w.r.t.x we have

2x-2+2y dy/dx-2dy/dx =0 So,x - 1+ydy/dx-dy/dx =0

So x-1+(y-1)dy/dx=0

dy/dx=1-x/y-1

differentiating again w.r.t.x,we get

1+(y-1)d²y/dx² +(dy/dx)^2=0

Now how to calculate required differential equation.

If my calculations of 1st and 2nd order derivative is wrong, reply me accordingly.

Differentiating both sides of $\displaystyle (x-h)^2+(y-k)^2=a^2$ we get $\displaystyle 2(x-h)+2(y-k)y'=0$ or equivalently $\displaystyle \boxed{(x-h)dx+(y-k)dy=0}$ (this is the solution)

Originally Posted by FernandoRevilla

Differentiating both sides of $\displaystyle (x-h)^2+(y-k)^2=a^2$ we get $\displaystyle 2(x-h)+2(y-k)y'=0$ or equivalently $\displaystyle \boxed{(x-h)dx+(y-k)dy=0}$ (this is the solution)

Sir,
I want the solution without arbitrary constants h,k. So I want to eliminate them by differentiation.
Okay,

Originally Posted by Vinod

Sir, I want the solution without arbitrary constants h,k. So I want to eliminate them by differentiation. Okay,

Why do you know that $\displaystyle h,k$ are arbitrary constants? Does the problem mention it? If this is the case, the problem should be stated: Find the differential equation of all circles in the plane.

Originally Posted by FernandoRevilla

Why do you know that $\displaystyle h,k$ are arbitrary constants? Does the problem mention it? If this is the case, the problem should be stated: Find the differential equation of all circles in the plane.

Sir,
You are correct. Original problem is as follows:
Find the differential equation of all circles having a fixed radius a.

Answer given in the study material is as below:
{1+ (dy/dx)^2}^3=a^2{d²y/dx²}^2

Hence give me a hint to arrive at the above answer.

Regards

Sir,
I found the differential equation of all circles having fixed radius a.First derivative of equation of these circles is
$\displaystyle \frac{dy}{dx}=\frac{-(x-h)}{(y-k)}$

Second derivative of equation of these circles is
d²y/dx²=[-(dy/dx)^2-1]/(y-k)

Differential equation of all circles having fixed radius a is given in post number #5 to this thread.

However thanks for your reply.