Solution: The given curve is \begin{equation}\label{eq:03Jan2025-1} x^2 + y^2 + 2ax + 2by + c = 0. \end{equation} Differentiating \eqref{eq:03Jan2025-1} with respect to $x$ thrice, we get \begin{gather} 2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} + 2a + 2b \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \label{eq:03Jan2025-2} \\[2ex] 1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 + y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \label{eq:03Jan2025-3} \\[2ex] 2 \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + d \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 0 \label{eq:03Jan2025-4}. \end{gather} In short, we can write, \begin{equation}\label{eq:03Jan2025-5} 3y'y'' + (y + b) y''' = 0. \end{equation}

Now we will find $b$ from the above equations. Using \eqref{eq:03Jan2025-3} and \eqref{eq:03Jan2025-4}, we eliminate $b$. More precisely, multiplying both side of \eqref{eq:03Jan2025-3} by $y'''$ and \eqref{eq:03Jan2025-5} by $y''$ and finally subtract them to get \[ \left[ 1 + \left( y' \right) ^2 \right] y''' - 3 y' \left( y'' \right) ^2 = 0. \] Thus, the corresponding differential equation will be \[ \textcolor{blue}{ \boxed{ \left[ 1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 \right] \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 3 \frac{\mathrm{d}y}{\mathrm{d}x} \left( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) ^2 = 0. } } \]