<s>[INST] How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$? I'm trying to find the orthogonal trajectories of the family of circles through the points $(1,1)$ and $(-1, -1)$. Now such a family can be given by an equation of the form $$ x^2 + y^2 + 2g(x-y) - 2 = 0, $$ where $g$ is a parameter. 
Now upon differentiation with respect to x, we obtain $$ 2x + 2y y^\prime + 2g (1 - y^\prime ) = 0, $$ where $y^\prime$ denotes the derivative of $y$ with respect to $x$. Upon dividing out by $2$, we arrive at $$ x + y y^\prime + g(1 - y^\prime ) = 0, $$ from which we get $$ g = \frac{x + y y^\prime}{y^\prime - 1}. $$ Putting this value of $g$ into the equation of the family of circles, we get $$ x^2 + y^2 +2 \frac{x + y y^\prime}{y^\prime - 1} ( x - y ) - 2 = 0, $$ so $$ (x^2 + y^2 -2 ) (y^\prime - 1 ) + 2 (x + y y^\prime ) (x - y)  = 0$$ or $$ (x^2 + y^2 - 2 + 2xy - 2y^2 ) y^\prime + (2x^2 - 2xy - x^2 - y^2 + 2) = 0  $$ or $$ ( x^2 + 2xy - y^2 - 2) y^\prime +  (x^2 - 2xy  - y^2 + 2) = 0, $$  from which we get $$ y^\prime = - \frac{ x^2 - 2xy  - y^2 + 2}{ x^2 + 2xy - y^2 - 2}. $$ Now for the orthogonal trajectories, we get $$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy  - y^2 + 2}. $$ How to solve this differential equation? 
 [/INST] Assuming your calculations are correct up to the last part, I begin with:
$$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy  - y^2 + 2}. $$
Observe that:
$$ y^\prime = \frac{1 + 2y/x - (y/x)^2 - 2/x^2}{1 - 2y/x  - (y/x)^2 + 2/x^2}. $$
Thus, let $v = y/x$ for which $y=vx$ and $y' = xv'+v$ and the given problem changes to:
$$ xv'+v = \frac{1 + 2v - v^2 - 2/x^2}{1 - 2v  - v^2 + 2/x^2}. $$
I suspect this can be solved. But, I'll stop here for now. This is a standard substitution for ODEs which can be written as a function of $y/x$ which reflects a certain symmetry, namely that $x$ and $y$ can be scaled by the same factor.
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