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Obtain the Differential equation with all Cocles pasing through the origin and \( (0,8) \)

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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.

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