Solution: We recall that the equation of a sphere with centre $(a,b,c)$ and radius $r$ is given by \[ (x-a)^2 + (y-b)^2 + (z - c)^2 = r^2. \] Since the sphere lies in the $xy$-plane, we have $c = 0$. Thus, the equation of the sphere becomes \[ (x-a)^2 + (y-b)^2 + z^2 = r^2. \]

Differentiating both sides with respect to $x$, we get \begin{align*} 2(x-a) + 2z \frac{\partial z}{\partial x} = 0 & \implies \frac{x - a}{z} = \frac{\partial z}{\partial x} \\ & \implies x - a = -z \frac{\partial z}{\partial x}. \end{align*} Similarly, differentiating with respect to $y$, we get \[ 2(y-b) + 2z \frac{\partial z}{\partial y} = 0 \implies y-b = -z\frac{\partial z}{\partial y}. \]

Substituting the values of $(x-a)$ and $(y-b)$ into the equation of sphere, we get \begin{align*} & \left( -z \frac{\partial z}{\partial x} \right) ^2 + \left( -z \frac{\partial z}{\partial y} \right) ^2 + z^2 = r^2 \\ \implies &z^2 \left( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} + 1 \right) = r^2. \end{align*} Thus, the required differential equation will be \[ \textcolor{blue}{ \boxed{ z^2 \left( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} + 1 \right) = r^2 } } \]