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)]=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)))`
D. None of the above

Answer

b

Explanation

Given , `(x-h)^(2) +(y-k)^(2) = a^(2) `
` rArr2(x-h)+2 (y-k)(dy)/(dx) =0`
` rArr (x-h) +(y-k) (dy)/(dx) =0" " ` … (ii)
Again , on differentiating , we get
` 1+ (y-k) (d^(2)y)/(dx^(2)) +((dy)/(dx))^(2)+((dy)/(dx))^(2)=0`
` rArr (y-k) = -(1+((dy)/(dx))^(2))/((d^(2)y)/(dx^(2)))`
Putting the value of `(y-k) ` in Eq. (ii) , we get
` x-h =-(y-k)(dy)/(dx)=([1+((dy)/(dx))^(2)](dy)/(dx))/((d^(2)y)/(dx^(2)))`
Putting the values of `(x-h) and (y-k)` 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)`
` rArr [1+((dy)/(dx))^(2)]^(2)[((dy)/(dx))^(2)+1] = a^(2) ((d^(2)y)/(dx^(2)))^(2)`
` rArr [ 1+ ((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`