Question

The differential equation of the systems of all circles of radius $r$ in the $xy$-plane is

• A. ${\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^2=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$
• B. ${\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^2=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^3$
• C. ${\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$
• D. ${\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^3$

Answer 1

The correct option is C

${\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$

Explanation for the correct option:

Finding the differential equation:

The general equation of circle is given by

${\left(\mathrm{x}-\mathrm{a}\right)}^2+{\left(\mathrm{y}-\mathrm{b}\right)}^2={\mathrm{r}}^2$

Now, on differentiating above equation we get

$$\begin{gathered} \Rightarrow 2\left(\mathrm{x}-\mathrm{a}\right)+2\left(\mathrm{y}-\mathrm{b}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \left(\mathrm{x}-\mathrm{a}\right)+2\left(\mathrm{y}-\mathrm{b}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \left(\mathrm{x}-\mathrm{a}\right)=-\left(\mathrm{y}-\mathrm{b}\right)\frac{\mathrm{dy}}{\mathrm{dx}} \end{gathered}$$

Again Differentiating above equation we get

$$\begin{gathered} \Rightarrow 1+\left(\mathrm{y}-\mathrm{b}\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=0 \\ \Rightarrow \left(\mathrm{y}-\mathrm{b}\right)=-\frac{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}\frac{\mathrm{dy}}{\mathrm{dx}} \end{gathered}$$

Now, substituting the value of $(x-a),(y-b)$in equation ${\left(\mathrm{x}-\mathrm{a}\right)}^2+{\left(\mathrm{y}-\mathrm{b}\right)}^2={\mathrm{r}}^2$ we get

${\left[\frac{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]}^2+{\left[\frac{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}{{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2}\right]}^2={\mathrm{r}}^2$

$$\begin{gathered} \Rightarrow {\left[1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]}^2\left[1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]=r^2\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right) \\ \Rightarrow {\left[1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]}^3=r^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2 \end{gathered}$$

Hence, option (C) is the correct answer.