Q49 / 120 · JEE Main 2007 · Mathematics
The differential equation of all circles passing through the origin and having their centres on the $x$-axis is
- A$x^2=y^2+x y \frac{d y}{d x}$
- B$x^2=y^2+3 x y \frac{d y}{d x}$
- Correct answerC$y^2=x^2+2 x y \frac{d y}{d x}$
- D$y^2=x^2-2 x y \frac{d y}{d x}$
Solution
General equation of all such circles is
$\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}=0$.
Differentiating, we get
$2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{g}=0$
$\therefore$ Desired equation is
$\begin{aligned}
& x^2+y^2+\left(-2 x-2 y \frac{d y}{d x}\right) x=0 \\
& \Rightarrow y^2=x^2+2 x y \frac{d y}{d x}.
\end{aligned}$
Hence, the correct answer is option: 'C'.