The differential equation of the family of curves of `x^(2)+y^(2)-2ay=0` where a is arbitary constant, is

Options

A. `(x^(2)+y^(2))(dy)/(dx)=2xy`
B. `2(x^(2)+y^(2))(dy)/(dx)=2xy`
C. `(x^(2)-y^(2))(dy)/(dx)=2xy`
D. `2(x^(2)-y^(2))(dy)/(dx)=2xy`

Answer

C

Explanation

We have,
`x^(2)+y^(2)-2ay=0" …(i)"`
Differentiating w.r. to x, we get
`2x+2y(dy)/(dx)-2a(dy)/(dx)=0rArra=(x+y(dy)/(dx))/((dy)/(dx))`
Substituting this value of a in (i), we get
`(x^(2)+y^(2))(dy)/(dx)-2y(x+y(dy)/(dx))=0rArr(x^(2)-y^(2))(dy)/(dx)=2xy`
This is the required differential equation.