Question

The differential equation of all circles which passes through the origin and whose centre lies on the X-axis is ______

विकल्प

• `(x^2 - y^2) - 2xydy/dx = 0`

• `(x^2 - y^2) + 2xydy/dx = 0`

• `(x^2 - y^2) - xydy/dx = 0`

• `(x^2 - y^2) + xydy/dx = 0`

Answer 1

The differential equation of all circles which passes through the origin and whose centre lies on the X-axis is `underline((x^2 - y^2) + 2xydy/dx = 0)`.

Explanation:

The system of circles which passes through origin and whose centre lies on the X-axis is

x² + y² - 2ax = 0 ................(i)

Differentiating w.r.t. x, we get

2x + 2y`dy/dx` - 2a = 0

⇒ `2a = 2x + 2ydy/dx` ......................(ii)

Substituting (ii) in (i), we get

`x^2 + y^2 - 2x^2 - 2xydy/dx = 0`

⇒ `(y^2 - x^2) - 2xydy/dx - 0`

⇒ `(x^2 - y^2) + 2xydy/dx = 0`