Solving PDE using characteristic method without polar coordinate. I'm trying to solve this PDE using method of characteristic without coordinate transformation :

$xu_x+yu_y = \sqrt{x^2+y^2}$ for $x^2+y^2>1$, $u(x,y)=x$ on $x^2+y^2 = 1$.

I only know how to use the method of characteristic for this kind of PDE such that initial condition is constant in one variable e.g., $u(0,y) = $ or $u(x,1) = $. I tried to mimick the technique to solve the above pde : Introducing new variable $(\xi,\eta)$ such that $\xi = x,\eta = y$ on $x^2+y^2 = 1$ and using chain rule,
$${\partial u\over\partial \eta} = {\partial u\over\partial x}{\partial x\over\partial\eta}+{\partial u\over\partial y}{\partial y\over\partial\eta} = \sqrt{x^2+y^2}$$
so ${\partial x\over\partial\eta}= x,{\partial y\over\partial \eta} = y$. So $x = \phi_1(\xi)e^{\eta}$ and $y = \phi_2(\xi)e^{\eta}$ ..? I'm stuck here and I'm not sure I'm in the right direction. Please help.

$$xu_x+yu_y = \sqrt{x^2+y^2}$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\sqrt{x^2+y^2}}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ :
$$y=c_1x$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{\sqrt{x^2+(c_1x)^2}}$
$$u-x\sqrt{1+c_1^2}=c_2$$
General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ :
$$u-x\sqrt{1+\left(\frac{y}{x}\right)^2}=F\left(\frac{y}{x}\right)$$
$F$ is an arbitrary function.
$$\boxed{u(x,y)=\sqrt{x^2+y^2}+F\left(\frac{y}{x}\right)}$$
Condition : $u(x,y)=x$ on $x^2+y^2 = 1$
$$\sqrt{x^2+(1-x^2)}+F\left(\frac{\sqrt{1-x^2}}{x}\right)=x$$
$$F\left(\frac{\sqrt{1-x^2}}{x}\right)=x-1$$
Let $X=\frac{\sqrt{1-x^2}}{x}\quad\implies\quad x=\frac{1}{1+X^2}$
$$F(X)=\frac{1}{\sqrt{1+X^2}}-1$$
Now the function $F(X)$ is determined. We put it into the above general solution where $X=\frac{y}{x}$
$$u(x,y)=\sqrt{x^2+y^2}+\frac{1}{\sqrt{1+\left(\frac{y}{x}\right)^2}}-1$$
$$\boxed{u(x,y)=\frac{x^2+y^2+x}{\sqrt{x^2+y^2}}-1}$$
