Solution: We will solve this by using the Lagrange's method. Recall that if \[ P(x,y,z) \frac{\partial x}{\partial z} + Q(x,y,z) \frac{\partial y}{\partial z} = R(x,y,z), \] then the auxiliary equations are \[ \frac{\mathrm{d} x}{P} = \frac{\mathrm{d} y}{Q} = \frac{\mathrm{d} z}{R}. \]

Here in the given problem, the auxiliary equations will be \[ \frac{\mathrm{d} x}{x} = \frac{\mathrm{d} y}{y} = \frac{\mathrm{d} z}{z}. \] We take \begin{align*} \frac{\mathrm{d} x}{x} = \frac{\mathrm{d} z}{z} & \implies \ln x = \ln z + \ln C_1 \\ & \implies \ln \left( \frac{x}{z} \right) = c \\ & \implies \frac{x}{z} = C_1. \end{align*}

Similarly, we have \[ \frac{\mathrm{d} y}{y} = \frac{\mathrm{d} z}{z} \implies \frac{y}{z} = C_2, \] and \[ \frac{\mathrm{d} x}{x} = \frac{\mathrm{d} y}{y} \implies \frac{x}{y} = C_3. \]

We can consider any two of the above equations. Let us consider \[ \frac{x}{z} = C_1, ~\text{ and }~\frac{y}{z}=C_2. \] Then, the general solution of the given PDE will be \[ f\left( \frac{x}{z},\frac{y}{z} \right) =0, \] where $f$ is an arbitrary function of two variables. This is equivalent to \[ \phi \left( \frac{x}{z} \right) = \frac{y}{z}, \] where $\phi $ is an arbitrary function of two variables.

Now we will use the initial conditions to determine the function. Note that if $x=t$, then \[ y = \sqrt{1-t^2},\text{ and } z=1. \] So, \begin{align*} \phi \left( \frac{x}{y} \right) = \frac{y}{z} & \implies \phi (t) = \sqrt{1-t^2} \\ & \implies \phi \left( \frac{x}{z} \right) = \sqrt{1-\left( \frac{x}{z} \right) ^2} \\ & \implies \frac{y}{z} = \sqrt{1-\frac{x^{2} }{z^{2} }} \\ & \implies \frac{y^2}{z^2}= 1-\frac{x^2}{z^2} \\ & \implies z^2 = x^2 + y^2\\ & \implies z = \sqrt{x^2 + y^2},~\text{since z>0}. \end{align*} Thus, the solution of the given Cauchy problem will be \[ \textcolor{blue}{\boxed{z = \sqrt{x^2 + y^2} }} \]