Find the solution for given PDE along unit circle

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The problem assigned to me is:

For the equation $$u_x^2 + u_y^2 = u^2$$ Find the solution when $u|_\Gamma$ = 1, where $\Gamma$ is unit circle at origin.

I was taught about method of characteristics so I tried to write characteristic equations as:

$$\begin{align} \frac{dx}{dt} &= 2p\\ \frac{dy}{dt} &= 2q\\ \frac{dz}{dt} &= 2z^2\\ \frac{dp}{dt} &= 2pz\\ \frac{dq}{dt} &= 2qz\\ \end{align}$$

I'm having a doubt here. Like why $\frac{dp}{dt}$ and $\frac{dq}{dt}$ aren't zero? They are supposed to be zero since we are considering them on characteristic curves right?

They being non-zero is making me wonder what impact does it have on the solution of given problem?

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By the symmetry of the problem, your solution will be radial. Let $$ v(r)=u(x,y),\qquad r^2=x^2+y^2. $$ The left hand side of your equation is the square of the gradient, which is nothing but $(v')^2$. So you have $$ (v')^2=v^2 $$ or $v'=\pm v$, with $v(1)=1$. You get two exponential solutions $$ v(r)= e^{\pm(r-1)}. $$ In terms of $(x,y)$, $$ u(x,y)= e^{\pm(\sqrt{x^2+y^2}-1)}. $$