Question

The differential equation of all circles passing through the origin and having their centre on the X-axis is

• A. \( x^2 = y^2 + xy\frac{dy}{dx} \)
• B. \( x^2 = y^2 + 3xy\frac{dy}{dx} \)
• C. \( y^2 = x^2 + 2xy\frac{dy}{dx} \)
• D. \( y^2 = x^2 - 2xy\frac{dy}{dx} \)

Hint:

Form equation $\longrightarrow$ differentiate $\longrightarrow$ eliminate parameter = standard DE approach.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A circle with its center on the X-axis and passing through the origin must have its center at some point \( (a, 0) \) and its radius equal to \( |a| \).
Step 2: Key Formula or Approach:
The general equation of such a circle is:
\[ (x - a)^2 + (y - 0)^2 = a^2 \]
Simplifying:
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
\[ x^2 + y^2 - 2ax = 0 \dots (i) \]
Step 3: Detailed Explanation:
Differentiating equation (i) with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} - 2a = 0 \]
\[ 2a = 2x + 2y \frac{dy}{dx} \]
\[ a = x + y \frac{dy}{dx} \]
Now, substitute the value of \( a \) back into equation (i):
\[ x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0 \]
\[ x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \]
\[ y^2 - x^2 - 2xy \frac{dy}{dx} = 0 \]
\[ y^2 = x^2 + 2xy \frac{dy}{dx} \]
Step 4: Final Answer:
The differential equation is \( y^2 = x^2 + 2xy \frac{dy}{dx} \).