Question:
The differential equation of all circles passing through the origin and having their centre on the X-axis is
The differential equation of all circles passing through the origin and having their centre on the X-axis is
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Form equation $\longrightarrow$ differentiate $\longrightarrow$ eliminate parameter = standard DE approach.
Updated On: Apr 22, 2026
- \( x^2 = y^2 + xy\frac{dy}{dx} \)
- \( x^2 = y^2 + 3xy\frac{dy}{dx} \)
- \( y^2 = x^2 + 2xy\frac{dy}{dx} \)
- \( y^2 = x^2 - 2xy\frac{dy}{dx} \)
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The Correct Option is D
Solution and Explanation
Concept:
Equation of circle with center on x-axis:
\[
(x-a)^2 + y^2 = a^2
\]
Step 1: Expand.
\[ x^2 -2ax + a^2 + y^2 = a^2 \Rightarrow x^2 + y^2 = 2ax \]
Step 2: Differentiate.
\[ 2x + 2y\frac{dy}{dx} = 2a \]
Step 3: Eliminate \(a\).
\[ a = \frac{x^2 + y^2}{2x} \] Substitute: \[ 2x + 2y\frac{dy}{dx} = \frac{x^2 + y^2}{x} \]
Step 4: Simplify.
\[ y^2 = x^2 - 2xy\frac{dy}{dx} \]
Step 1: Expand.
\[ x^2 -2ax + a^2 + y^2 = a^2 \Rightarrow x^2 + y^2 = 2ax \]
Step 2: Differentiate.
\[ 2x + 2y\frac{dy}{dx} = 2a \]
Step 3: Eliminate \(a\).
\[ a = \frac{x^2 + y^2}{2x} \] Substitute: \[ 2x + 2y\frac{dy}{dx} = \frac{x^2 + y^2}{x} \]
Step 4: Simplify.
\[ y^2 = x^2 - 2xy\frac{dy}{dx} \]
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