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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?

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Let $u = y+x$, $v = y-x$, we have

$$\begin{align} u' = y'+1 = & \frac{2x^2-2y^2}{x^2 - 2xy - y^2+2} = \frac{-2uv}{x^2-2xy-y^2+2}\\ v' = y'-1 = & \frac{4xy-4}{x^2 -2xy- y^2+22} = \frac{u^2 - v^2-4}{x^2-2xy-y^2+2}\\ \end{align}$$

From this, we get $$u^2 \left( \frac{v^2}{u} + u +\frac{4}{u} \right)' = u^2\left(\frac{v^2}{u}\right)' + (u^2 - 4)u' = 2uvv' + (u^2 - v^2 - 4)u' = 0$$ and hence for some integration constant $K$, one has $$\frac{v^2}{u} + u +\frac{4}{u} = 2K \;\;\iff\;\; v^2 + u^2 + 4 = 2K u \;\;\iff\;\; x^2 + y^2 + 2 = K(x+y) $$ i.e the orthogonal trajectories is another family of circles.

Update

There is a pure geometric argument why the orthogonal trajectories are circles.

Let $p = (-1,-1), q = (1,1)$. Let $\mathscr{C}$ be a circle centered at $p$ with radius $|pq| = 2\sqrt{2}$. Consider the inversion $\mu$ with respect to circle $\mathscr{C}$, we have:

$$\mu(p) = \infty\quad\text{ and }\quad \mu(q) = q.$$

This means under $\mu$, the family of circles passing through $p$ and $q$ get mapped to the family of straight lines passing through $q$. The orthogonal trajectories of this family of straight lines is the family of circles centered at $q$.

It is known that inversion is conformal. i.e. preserve angles and hence map orthogonal trajectories to orthogonal trajectories. This means the orthogonal trajectories of the family of circles passing through $p$ and $q$ are the inverse image of $\mu$ of the family of circles centered at $q$.

It is also known that inversion $\mu$ map circles not passing through $q$ to circles. As a result, the family of orthogonal trajectories we seek are circles themselves.

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  • $\begingroup$ achille hui, fantastic work. $\endgroup$ – Saaqib Mahmood Oct 14 '13 at 14:11
  • $\begingroup$ I just dont see how you go from this: $$\begin{align} u' = y'+1 = & \frac{2x^2-2y^2}{x^2 - 2xy - y^2+2} = \frac{-2uv}{x^2-2xy-y^2+2}\\ v' = y'-1 = & \frac{4xy-4}{x^2 -2xy- y^2+22} = \frac{u^2 - v^2-4}{x^2-2xy-y^2+2}\\ \end{align}$$ to this: $$u^2 \left( \frac{v^2}{u} + u +\frac{4}{u} \right)' = u^2\left(\frac{v^2}{u}\right)' + (u^2 - 4)u' = 2uvv' + (u^2 - v^2 - 4)u' = 0$$ Sorry, but someone needs to explain this a little more. (PS I get the substitutions, I just don't see the next step!) $\endgroup$ – emjay Oct 28 '15 at 2:18
  • $\begingroup$ @emjay The are two phases in constructing this proof. In the discovering phase, I start from the observation on RHS that $2uvv' + (u^2-v^2-4)u' = 0$, working backwards from left to right, combine term until you get something nice on LHS. When I get to second phase of write the proof down. I pull out the expression on LHS from thin air, reverse the steps and demonstrate RHS is true. When one write down a proof, it is very common to skip the motivations and minimize the number of steps to show the correctness of the final statement. $\endgroup$ – achille hui Oct 28 '15 at 3:28
  • $\begingroup$ Thank you, I just worked through the solution using your observations. $\endgroup$ – emjay Oct 28 '15 at 14:58
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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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  • $\begingroup$ James S. Cook, as far as my knowledge goes, we can use this substitution only if the right hand side is purely a function of $y/x$, whereeas in this case there're terms involving $1/x^2$. I'm afraid the substitution you suggested might not work. $\endgroup$ – Saaqib Mahmood Oct 13 '13 at 19:18
  • $\begingroup$ SaaqibMahmuud you may well be correct, in any event, I see a linear substitution has tamed the beast. Nice work @achille hui $\endgroup$ – James S. Cook Oct 13 '13 at 19:53

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