Let the equation of circle is x2 + y2 + 2fx + 2gy + c = 0 .....(1)
Since, given that circle is passing through origin.
∴ c = 0
Also circle is passing through point (0, 8).
∴ 64 + 16g = 0 (∵ c = 0)
⇒ g = -64/16 = -4
Hence, the equation of given circle is x2 + y2 + 2fx - 8y = 0 ......(2)
Since, equation (2) contains one arbitrary constant. Therefore, we differentiate equation (2) one time w.r.t. x and hide the arbitrary constant to get differential equation.
Differentiate equation (2) w.r.t. x, we get
2y \(\frac{dy}{dx}\) + 2x + 2f - 8 \(\frac{dy}{dx}\) = 0
⇒ f = (4 - y) \(\frac{dy}{dx}\) - x
Put the value of f in equation (2), we get
x2 + y2 + 2x(4 - y) \(\frac{dy}{dx}\) - 2x2 - 8y = 0
⇒ \(\frac{dy}{dx}=\frac{x^2-y^2+8y}{2x(4-y)}\) .......(3)
Equation (3) represents differential equation of given circle.