Question

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

• A. \( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \)
• B. \( \frac{dy}{dx} = \frac{2xy}{x^2 + y^2} \)
• C. \( \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \)
• D. None of these

Hint:

Form equation → differentiate → eliminate parameter.

Answer 1

The Correct Option is C

Concept: Circle with centre on Y-axis: \[ x^2 + (y - a)^2 = a^2 \]

Step 1: Simplify equation \[ x^2 + y^2 - 2ay + a^2 = a^2 \Rightarrow x^2 + y^2 - 2ay = 0 \]

Step 2: Differentiate \[ 2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0 \] \[ 2x + 2(y-a)\frac{dy}{dx} = 0 \] \[ (y-a)\frac{dy}{dx} = -x \]

Step 3: Eliminate \(a\) From original equation: \[ x^2 + y^2 = 2ay \Rightarrow a = \frac{x^2 + y^2}{2y} \] \[ y - a = y - \frac{x^2 + y^2}{2y} = \frac{2y^2 - x^2 - y^2}{2y} = \frac{y^2 - x^2}{2y} \]

Step 4: Substitute \[ \frac{y^2 - x^2}{2y}\frac{dy}{dx} = -x \] \[ \frac{dy}{dx} = \frac{-2xy}{y^2 - x^2} = \frac{2xy}{x^2 - y^2} \] Conclusion \[ {\frac{dy}{dx} = \frac{2xy}{x^2 - y^2}} \]