Manipal MET2023MathematicsCircle
The differential equation of all circles which pass through the origin and whose centres lies on Y -axis is
Options
- Ad y d x = x y x^2+y^2
- Bd y d x = 2 x y x^2+y^2
- Cd y d x = 2 x y x^2-y^2
- DNone of these
Correct answer
C. d y d x = 2 x y x^2-y^2
Step-by-step solution
The equation of family of circles which passess through the origin and whose centrer lies on Y -axis is (x-0)^2+(y-k)^2=k^2 aligned & x^2+y^2+k^2-2 y k=k^2 & x^2+y^2-2 y k=0 aligned where, k is parameter. On differentiating w.r.t . x , we get 2 x+2 y d y d x -2 d y d x k=0 array ll & x+(y-k) d y d x =0 & x+ y- x^2+y^2 2 y d y d x =0 array array ll & x+ 1 2 y (y^2-x^2 ) d y d x =0 & 2 x y= (x^2-y^2 ) d y d x array d y d x = 2 x y x^2-y^2 , which is required equation.