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

Step 1: Understanding the Concept:
A circle passing through the origin \((0,0)\) with center on the Y-axis at \((0,k)\) must have radius \(|k|\). We define the general equation and differentiate to eliminate the arbitrary constant \(k\).
Step 3: Detailed Explanation:
1. General Equation: \(x^{2} + (y - k)^{2} = k^{2}\).
Expand: \(x^{2} + y^{2} - 2yk + k^{2} = k^{2} \implies x^{2} + y^{2} = 2yk\).
2. Differentiate with respect to \(x\):
\[ 2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx} \]
\[ x + y \frac{dy}{dx} = k \frac{dy}{dx} \implies k = \frac{x + y y'}{y'} \]
3. Substitute the value of \(k\) back into the expansion \(x^{2} + y^{2} = 2yk\):
\[ x^{2} + y^{2} = 2y \left( \frac{x + y y'}{y'} \right) \]
4. Multiply both sides by \(y'\):
\[ (x^{2} + y^{2}) y' = 2xy + 2y^{2} y' \]
\[ x^{2} y' + y^{2} y' - 2y^{2} y' = 2xy \]
\[ (x^{2} - y^{2}) y' = 2xy \]
\[ \frac{dy}{dx} = \frac{2xy}{x^{2} - y^{2}} \]
Step 4: Final Answer:
The required differential equation is \(\frac{dy}{dx} = \frac{2xy}{x^{2} - y^{2}}\).