Solve $\{(\frac{dy}{dx})^2+1\}(x-y)^2=(x+y\frac{dy}{dx})^2$ The question is as follows:

Solve $\{(\frac{dy}{dx})^2+1\}(x-y)^2=(x+y\frac{dy}{dx})^2$ .

Can we reduce it to Clairaut's form by taking appropriate substitution? Please help.
 A: Why do you think that it can be reduced to a Clairaut's form of ODE in order to solve it ?
$$\{(\frac{dy}{dx})^2+1\}(1-\frac{y}{x})^2=(1+\frac{y}{x}\frac{dy}{dx})^2$$
Obviously this is an ODE of the homogeneous kind. The usual way to solve it is a change of function : $u=\frac{y}{x}$
$\frac{dy}{dx}=u+x\frac{du}{dx}$
Bringing it into the ODE and solving it for $\frac{du}{dx}$ leads to :
$$\frac{du}{dx}=\frac{ -2u^2 \pm (u-1)\sqrt{2u} }{x(2u-1)}$$
$$\int \frac{dx}{x}=\int{\frac{2u-1}{-2u^2 \pm (u-1)\sqrt{2u}}}$$
After integration and exponentiation :
$$x=\frac{c}{u+1\pm \sqrt{2u}}$$
Solving it for $u$ gives :
$$u=\frac{c\pm\sqrt{-x^2+2cx}}{x}$$
and finally the solution :
$$y=c\pm\sqrt{-x^2+2cx}$$
or, expressed on implicit form :
$$(y-c)^2+(x-c)^2=c^2$$
A: HINT:
$$  2 \sqrt {(1 +{\frac{dy}{dx} })^2 } (x-y) = \frac{d(x^2+y^2)}{dx} $$
Attempt some simplifications with: 
$ x= r \cos t, y = r \sin t. $
A: Its a bit messy but, here is my solution:


*

*set $p= \frac{dy}{dx}=y'$
$(p^2+1)(x-y)^2 = (x+yp)^2$

*take square root of everyting
$\sqrt{(p^2+1)}|(x-y)| = |(x+yp)|$

*take one case when $x-y > 0$ and  $x +yp > 0$, get
$\sqrt{(p^2+1)}(x-y) = (x+yp)$

*expand left side, regroup
$x\sqrt{(p^2+1)}- x = yp + y\sqrt{(p^2+1)}$ 
$x(\sqrt{(p^2+1)}- 1) = y(p + \sqrt{(p^2+1)})$

*now you get form $y= F(p)x $
$x \frac{\sqrt{(p^2+1)}- 1}{p + \sqrt{(p^2+1)}} = y$

*Differentiate that form to get:
$y' = p = F'(p)p'x + F(p)$

*Now you got rid of y, after regrouping
$\frac{F'(p) p'}{p-F(p)} = 1/x$

*insert your $F(p)$ and $F'(p)$, and integrate, it's a mess without clever substitutions but I just plugged in everything in Mathematica, now you get :
$\frac{p+\sqrt{p^2+1}}{\sqrt{p^2+1}}=c_1 x$

*solving for $p$
$p=\pm \frac{i (c_1 x-1)}{\sqrt{c_1} \sqrt{x} \sqrt{c_1 x-2}}$

*now integrate that, get:
$y= \pm \frac{i \sqrt{x} \sqrt{c_1 x-2}}{\sqrt{c_1}} + c_2$

*pluging in 1. you get one possible solution for constants that work for all x 
$c_1 =c_2 =1 $

*so one final solution to (1) is 
$y= \pm i*\sqrt{x(x-2)} + 1$
you can expolore in Mathematica other solutions by plugging 10 into 1 (NOT 2) and making sure it will all be 0. Im sure there are smarter solutions to this, at least I hope it helps.
