Find the differential equation of all the circles which pass through the origin and whose centres lie on y-axis.

Options

A. `(dy)/(dx)=(xy)/(x^(2)+y^(2))`
B. `(dy)/(dx)=(2xy)/(x^(2)+y^(2))`
C. ` (dy)/(dx)= (2xy)/(x^(2)-y^(2))`
D. None of these

Answer

c

Explanation

The equation of family of ciircles which passes through the origin and whose centres lie on Y - axis is ,
`(x-0)^(2) + ( y-k)^(2) = k^(2)`
` rArr x^(2) + y^(2) + k^(2) - 2 y k = k^(2))`
` rArr x^(2) + y^(2) + k^(2) - 2yk = k^(2)`
` rArr x^(2) + y^(2) - 2yk = 0 " "` ...(i)
where k is parameter
On differentiate w.r.t x, we get
` 2x + 2y (dy)/(dx) - 2 (dy)/(dx) k = 0 `
` rArr x + (y-k) (dy)/(dx) = 0 `
` rArr x + { y - (x^(2) +y^(2))/(2y)}(dy)/(dx) = 0 ` [ from Eq. (i)]
` rArr x + 1/(2y) (y^(2)-x^(2)) (dy)/(dx) = 0 `
` rArr 2xy = (x^(2) - y^(2)) (dy)/(dx) rArr (dy)/(dx) = (2xy)/(x^(2)-y^(2))`
which is the required eqution .