The differential equation whose solution is `(x-h)^2+ (y-k)^2=a^2` is (a is a constant)

Options

A. `[1+((dy)/(dx))^(2)]^(3)=a^(2)(d^(2)y)/(dx^(2))`
B. `[1+((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`
C. `[1+((dy)/(dx))]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`
D. None of the above

Answer

B

Explanation

Given, `(x-h)^(2)+(y-k)^(2)=a^(2)" (i)"`
`implies2(x-h)+2(y-k)(dy)/(dx)=0`
`implies(x-h)+(y-k)(dy)/(dx)=0" "...(ii)`
Again differentiating
`(y-k)=-(1+((dy)/(dx))^(2))/(d^(2)y//dx^(2))`
Putting in Eq. (ii), we get
`x-h =-(y-k)(dy)/(dx)=([1+((dy)/(dx))^(2)](dy)/(dx))/((d^(2)y)/(dx^(2)))`
Putting in Eq. (i), we get
`([1+((dy)/(dx))^(2)]^(2)((dy)/(dx))^(2))/(((d^(2)y)/(dx^(2)))^(2))+([1+((dy)/(dx))^(2)]^(2))/(((d^(2)y)/(dx^(2)))^(2))=a^(2)`
`implies[1+((dy)/(dx))^(2)]^(2)[((dy)/(dx))^(2)+1]=a^(2)((d^(2)y)/(dx^(2)))^(2)`
`implies[1+((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`