Correct Answer - B
We have,
`x^(2)+y^(2)-ay=0`
`rArr" "2x+2y(dy)/(dx)-a(dy)/(dx)=0rArra=(2(x+y(dy)/(dx)))/((dy)/(dx))`
Substituting the value of a in (i), we obtain
`x^(2)+y^(2)-2y((x+y(dy)/(dx))/((dy)/(dx)))=0rArr(x^(2)-y^(2))(dy)/(dx)-2xy=0`
This is differential equaiton of the family of circles given in
(i). The differential equation representing the orthogonal trjectories is
`-(x^(2)-y^(2))(dx)/(dy)-2xy=0`
`rArr" "(dy)/(dx)=-((x^(2)-y^(2))/(2xy))`
`rArr" "2xy dy-y^(2)dx=-x^(2)dx`
`rArr" "(xd(y^(2))-y^(2)dx)/(x^(2))=-dxrArrd((y^(2))/(x))=-dx`
On integrating, we get
`(y^(2))/(x)=-x+CrArry^(2)+x^(2)=Cx`
`ul("REMARK")` If the orthogonal trajectories of a family of curves is same as the given family of curves, then the given family of curves is called self- orthogonal.
