Differential equations of first order but of higher degree Solve $(x-py)(px-y)=2p$ where $p=\frac{dy}{dx}$
My attempt:
I started by making it a quadratic in $p$ but after a point a could not proceed further:
$$p^2-\frac{p(x^2+y^2-2)}{xy} + 1 = 0$$
$$p = (x^2+y^2-2)\pm\frac{(x^2-y^2)^2-4(x^2+y^2-1)^{1/2}}{2xy}$$
Now I don't know how to proceed further.

We are trying to solve 
$$\tag 1 (px-y)(x-py)=2p , ~\text{where}~p=\frac{dy}{dx}$$
Continuing on the path used leads to something unsolvable, so let's try another approach.
Let $x^2 = X \implies 2 x dx = dX, y^2 = Y \implies 2 y dy = dY$, so
$$P = \dfrac{dY}{dX} = \dfrac{2 y dy}{2 x dx} = \dfrac{y}{x} p \implies p = \dfrac{x}{y}P$$
Substituting into $(1)$
$$\left(\left(\dfrac{x}{y}  P\right) x - y\right)\left(x - \left(\dfrac{x}{y} P\right) y \right) = 2 \dfrac{x}{y}P$$
Simplifying
$$\left( \dfrac{x^2}{y}P - y\right) (x(1-P))= 2 \dfrac{x}{y}P$$
Solving for $y$
$$y^2 = P x^2 - 2 \dfrac{P}{1-P} = P X - 2 \dfrac{P}{1-P}$$
This is a Clairaut's Equation of the form $y = px + f(p)$, so we can write the solution as 
$$y^2 = c x^2 -2 \dfrac{c}{1-c}$$
