Q73 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
The differential equation of family of circles whose centre lie on $\mathrm{X}$-axis is
- A$\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0$
- Correct answerB$y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2+1=0$
- C$y\left(\frac{d^2 y}{d x^2}\right)-\left(\frac{d y}{d x}\right)^2-1=0$
- D$y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2-1=0$
Solution
Let $(h, 0)$ be the centre of the circle and ' $r$ ' be the radius.
$$
\therefore(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-0)^2=\mathrm{r}^2 \Rightarrow(\mathrm{x}-\mathrm{h})^2+\mathrm{y}^2=\mathrm{r}^2
$$
Differentiating w.r.t. $\mathrm{x}$, we get
$$
\therefore 2(\mathrm{x}-\mathrm{h})(1)+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad \Rightarrow \mathrm{h}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}
$$
Substituting value of ' $h$ ' in eq. (1), we get
$$
y^2\left(\frac{d y}{d x}\right)^2+y^2=r^2 \Rightarrow y^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=r^2
$$
Differentiating w.r.t. $\mathrm{x}$, we get
$$
\begin{aligned}
& y^2\left(2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}\right)+\left[1+\left(\frac{d y}{d x}\right)^2\right] 2 y \frac{d y}{d x}=0 \\
& \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.