The differential equation of all circles which pass through the origin and whose centres lie on $y$ -axis is

(x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0

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(x^{2} - y^{2}) \frac{d y}{d x} + 2 x y = 0

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(x^{2} - y^{2}) \frac{d y}{d x} - x y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x^{2} - y^{2}) \frac{d y}{d x} + x y = 0

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Solution

The correct option is B $(x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0$
Equation of a circle is
$x^{2} + (y - a)^{2} = a^{2}$ ...(i)
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{d y}{d x} = \frac{2 x y}{x^{2} - y^{2}}$
$\Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0$

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