Q23 / 50 · Manipal (MET) 2023 · Mathematics
The differential equation of all circles which pass through the origin and whose centres lies on $Y$-axis is
- A$\frac{d y}{d x}=\frac{x y}{x^2+y^2}$
- B$\frac{d y}{d x}=\frac{2 x y}{x^2+y^2}$
- Correct answerC$\frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$
- DNone of these
Solution
The equation of family of circles which passess through the origin and whose centrer lies on $Y$-axis is $(x-0)^2+(y-k)^2=k^2$
$\begin{aligned} & \Rightarrow x^2+y^2+k^2-2 y k=k^2 \\ & \Rightarrow x^2+y^2-2 y k=0\end{aligned}$
where, $k$ is parameter.
On differentiating w.r.t . $x$, we get
$2 x+2 y \frac{d y}{d x}-2 \frac{d y}{d x} k=0$
$\begin{array}{ll}\Rightarrow & x+(y-k) \frac{d y}{d x}=0 \\ \Rightarrow & x+\left\{y-\frac{x^2+y^2}{2 y}\right\} \frac{d y}{d x}=0\end{array}$
$\begin{array}{ll}\Rightarrow & x+\frac{1}{2 y}\left(y^2-x^2\right) \frac{d y}{d x}=0 \\ \Rightarrow & 2 x y=\left(x^2-y^2\right) \frac{d y}{d x}\end{array}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$,
which is required equation.
Hence, the correct answer is option: 'C'.