Question

find the differential equation of all circles which pass through the origin and whose center lies on xaxis

Answer 1

dy/dx = (y² - x²)/2xy   or y² = x² +  2xy(dy/dx) is the differential equation of all circles which pass through the origin and whose center lies on x axis

Step-by-step explanation:

center lies on x axis

let say cenetr = ( a , 0)

Radius = a  as circle passes through origin

=> (x - a)² + (y - 0)² = a²

=> x² + a²  - 2ax  + y² = a²

=>  x²  - 2ax  + y² =  0

=> 2ax = x² + y²

=> 2a  = (x² + y²)/x

=> 2a  =  x   +  y²/x

Differentiating wrt x

=> 0  = 1   + 2y(dy/dx)/x    - y²/x²

=> 2y(dy/dx)/x   =  y²/x² - 1

=> 2xy(dy/dx) = y² - x²

=> dy/dx = (y² - x²)/2xy

or y² = x² +  2xy(dy/dx)

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