Question

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

Answer 1

Equation of circle which passes through the origin and whose centre lies on y-axis is

(x – 0)² + (y – a)² = a²

⇒ x² + y² + a² – 2ay = a²

⇒ x² + y² – 2ay = 0   ......(i)

Differentiating both sides w.r.t. x we get

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

⇒ `x + y "dy"/"dx" - "a" * "dy"/"dx"` = 0

⇒ `x + (y - "a") * "dy"/"dx"` = 0

`y - "a" = x/("dy"/"dx")`

a = `y + (-x)/("dy"/"dx")`

Putting the value of a in equation (i), we get

`x^2 + y^2 - 2(y + x/("dy"/"dx"))y` = 0

⇒ `x^2 + y^2 - 2y^2 - (2xy)/("dy"/"dx")` = 0

⇒ `x^2 - y^2 = (2xy)/("dy"/"dx")`

∴ `(x^2 - y^2) "dy"/"dx" - 2xy` = 0

Hence, the required differential equation is `(x^2 - y^2) "dy"/"dx" - 2xy` = 0