Consider the wave equation in two spatial dimensions
$$\tag{1} \partial_{tt}u-c^2(\partial_{xx}u+\partial_{yy}u)=0 $$
I want to show that an implicit surface $\phi(x,y,t)=0$ is a characteristic surface of (1) if
$$\tag{2} (\partial_t\phi)^2-c^2\left[(\partial_x\phi)^2+(\partial_y\phi)^2\right]=0 $$
Question: How to derive (2)?
Details: We suppose that Cauchy data: $u\big|_\phi$ (the value of $u$ everywhere on $\phi$) and $\partial_\phi u \big|_\phi$ (the normal derivative of $u$ to $\phi$) have been supplied on the surface $\phi$. We will call $\phi$ a characteristic surface if the quantity $\partial_{\phi \phi} u \big|_\phi$ cannot be computed from the Cauchy data and the differential equation (1).
Context: The example, as well as (2) is from these notes section 8.3.
What I think: I understand the simpler example (section 8.2) for a PDE of two variables
$$ a \partial_{xx}u+b\partial_{xy}u+c\partial_{yy}u=d $$
with an initial condition; $u$ and the normal derivative of $u$, given along the parametric curve $(x(s),y(s))$. In this case one forms a system of equations for the unknowns $\partial_{xx}u, \ \partial_{xy}u, \ \partial_{yy}u$, which fails to have a unique solution if
$$\tag{3} ay'(s)^2-2bx'(s)y'(s)+cx'(s)^2=0 $$
It is not obvious to me how an analogous argument will yield (2).
Consider the change of co-ordinates from $x,y,t$ to $\phi,r,s$. We want to rewrite (1) in the new co-ordinates. When we do, we will find; schematically,
$$\tag{1b} F \ \partial_{\phi\phi}u+\text{lower/other derivatives}=0 $$
Where $F=F(\partial_x\phi,\partial_y\phi,\partial_t\phi)$. The surfaces $\phi(x,y,t)=\text{constant}$ will be characteristic surfaces if $F=0$ so that $\partial_{\phi\phi}u$ cannot be obtained from (1b). Using the chain rule and picking out only the coefficients of $\partial_{\phi\phi}u$ we find directly that $F=\text{LHS of (2)}$.