Non-linear partial differential equation of order 1 Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $\displaystyle p = \frac{\partial z}{\partial x}$, $\,\displaystyle q = \frac{\partial z}{\partial y}$
$f_p=2q+2x\\ f_q=2p+2x\\ f_z=0\\ f_x=2q+2x\\  f_y=2p+2y$
Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.

Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
Adding first two and last two fractions
$$\frac{dp+dq}{2p+2y+2q+2x}=\frac{dx+dy}{-2q-2y-2p-2x}$$
This implies, $a$ being an arbitrary constant $$p+x+q+y=a \tag{1}$$
Now the equation can be rewritten as $$2(p+x)(q+y)+(x-y)^2=0\tag{2}$$
Using $1,2$ $$(p+x)-(q+y)=\sqrt{a^2-2(x-y)^2}$$
Solving for $p+x$ and $q+y$ ,
$$\begin{align}  
p+x &=  \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q+y &=  \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
This gives
$$\begin{align}  
p&= -x+ \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q &= -y+ \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
$$dz=pdx+qdy$$
Substituting and integrating gives $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y) +\frac12\int\sqrt{a^2-2(x-y)^2}d(x-y)$$
So the solution is $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y)+\dfrac{a^2\ln\left(\left|a\sqrt{2(x-y)^2+a^2}+\sqrt{2}\left|a\right|(x-y)\right|\right)}{2^\frac{5}{2}}+\dfrac{a(x-y)\sqrt{\frac{2(x-y)^2}{a^2}+1}}{2^2}+b$$
