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6. ⇒ (MHT CET 2021 20th September Morning Shift )

The differential equation of all circles which pass through the origin and whose centre lie on Y-axis is

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

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

C. $(x^{2} + y^{2}) \frac{d y}{d x} + 2 x y = 0$

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

Correct Option is (A)

MHT CET 2021 20th September Morning Shift Mathematics - Differential Equations Question 37 English Explanation

Let (0, K) be the centre of the circle.

Since the circle passes through origin, we write

$\begin{aligned} (x - 0)^{2} + (y - K)^{2} = K^{2} \\ ∴ x^{2} + y^{2} - 2 K y + K^{2} = K^{2} \\ ∴ x^{2} + y^{2} - 2 K y = 0 . . . (1) \end{aligned}$

Differentiating w.r.t. $x$ , we get

$2 x + 2 y \frac{d y}{d x} - 2 K \frac{d y}{d x} = 0$

$∴ K = \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}}$ and substituting value of K in (1), we get

$\begin{aligned} x^{2} + y^{2} - 2 (\frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}}) y = 0 \\ ∴ & (x^{2} + y^{2}) \frac{d y}{d x} - 2 x y - 2 y^{2} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0 \end{aligned}$

⇒Defferential Equation