Q92 / 150 · MHT CET 2011 · Mathematics
The differential equation whose solution is $(x-h)^{2}+(y-k)^{2}=a^{2}(a$ is a constant $)$, is
- A$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2} \frac{d^{2} y}{d x^{2}}$
- Correct answerB$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
- C$\left[1+\left(\frac{d y}{d x}\right)\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
- DNone of the above
Solution
Given, $(x-h)^{2}+(y-k)^{2}=a^{2} \quad$...(i)
$$
\begin{array}{lr}
\Rightarrow & 2(x-h)+2(y-k) \frac{d y}{d x}=0 \\
\Rightarrow & \quad(x-h)+(y-k) \frac{d y}{d x}=0
\end{array}
$$
Again differentiating
$$
(y-k)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{d^{2} y / d x^{2}}
$$
Putting in Eq. (ii), we get
$$
\begin{array}{l}
x-h=-(y-k) \frac{d y}{d x} \\
=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d x^{2}}}
\end{array}
$$
Putting in Eq. (i), we get
$$
\begin{array}{l}
\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}} \\
\Rightarrow \quad\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2} \\
+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=a^{2} \\
\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}
\end{array}
$$
Hence, the correct answer is option: 'B'.