1.

The orthogonal trajectories of the family of circles given by `x^2 + y^2 - 2ay = 0`, isA. `x^(2)+y^(2)-2kx=0`B. `x^(2)+y^(2)-2ky=0`C. `x^(2)+y^(2)-2k_(1)x-2k_(2)y=0`D. none of these

Answer» Correct Answer - A
The equation of the family of circles is
`x^(2)+y^(2)-2ay=0" …(i)"`
Differentiating w.r.t. x, we get
`2x+2y(dy)/(Dx)-2a(dy)/(dx)=0rArra=(x+y(dy)/(dx))/((dy)/(dx))`
Putting this value of a in (i), we get
`x^(2)+y^(2)-2y{(x+y(dy)/(dx))/((dy)/(dx))}=0rArrx^(2)-y^(2)-2xy(dx)/(dy)=0" ...(ii)"`
This is the differential equation of the family of circles givne by (i).
The differential equation representing the family of orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)by-(dx)/(dy)` in (ii). So, the differential equation of the orthogoanl trajectories is
`x^(2)-y^(2)+2xy(dy)/(dx)=0`
`rArr" "(x^(2)-y^(2))dx+2xydy=0`
`rArr" "(xd(y^(2))-y^(2))/(x^(2))=-dx`
`rArr" "d((y^(2))/(x))=-dx`
`rArr" "(y^(2))/(x)=-x+2krArrx^(2)+y^(2)-2kx=0`
This is the required family of orthogonal trajectories.


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