Finding a differential equation orthogonal to a family of curves

Question

The question is: Consider the family $F$ of circles in the $xy$ plane, $(x-c)^2+y^2=c^2$ tangent to the $y$ axis at the origin. Find a differential equation that is satisfied by the family of curves orthogonal to $F$.

My thinking: Since the implicit equation represents the level sets of the function $$ f(x,y)=c^2=(x-c)^2+y^2 $$ The gradient of the function $f$ will be perpendicular to its level sets, and therefore orthogonal to the family of curves $F$. This yields $$ \nabla f(x,y)=(0,0)=(2x-2c,2y)\Rightarrow \left(x-\frac{x^2+y^2}{2x},y\right)=(0,0)\\ \Rightarrow \left(\frac{x^2-y^2}{2x},y\right)=(0,0) $$ So we have in differential form $$ \frac{x^2-y^2}{2x}dx+ydy=0\Rightarrow \frac{y^2-x^2}{2xy}=\frac{dy}{dx} $$ But the answer is the negative reciprocal, or perpendicular vector to this one. Why? I assume my reasoning was flawed in the first step, when i took the gradient of $f$ to be perpindicular to the family $F$, but I don't see why.

Answer 1

$$(x-c)^2+y^2=c^2$$ or : $$y^2+x^2-2cx=0 \quad\to\quad 2c=\frac{y^2+x^2}{x}=\frac{y^2}{x}+x$$ The differential equation of this family of circles is obtained by differentiation : $$dc=0=2\frac{y}{x}dy-\frac{y^2}{x^2}dx+dx$$ $$2\frac{y}{x}dy=\left(\frac{y^2-x^2}{x^2}\right)dx$$ $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$ This is the equation that you found. But, this differential equation is for the family of circles, not for the family of orthogonal curves.

The differential equation of the family of orthogonal curves is : $$-\frac{dx}{dy}=\frac{y^2-x^2}{2xy}$$ $$\frac{dy}{dx}=-\frac{2xy}{y^2-x^2}$$

Answer 2

The differential of the equation is $2(x-c)\;dx+2y \;dy=0$ and describes the tangents of a curve at a point $(x,y)$ and with $c=c(x,y)$ verifying $2cx =x^2+y^2$. By exchanging coefficients and changing sign on one (so as to get the orthogonal vector) we get $-2y \; dx + 2(x-c) \; dy=0$ or $0= -2yx \; dx + (2x^2-2xc) \;dy = -2yx \; dx + (x^2-y^2)\; dy$. Thus,

$$2yx \; dx + (y^2-x^2)\; dy =0$$ which describes the normal to a curve at $(x,y)$ and whence integrates to the wanted orthogonal curves.

The ODE is $\frac{dy}{dx}=\frac{2yx}{x^2-y^2}$, but is somewhat singular.

The above differential form integrates to a familiy of circles centered on the $y$-axis and going through the origin. It is dual to the given family and are also called "Apollonian circles".

More precisely, $x^2+(y-b)^2=b^2$ has as differential $2x\; dx+2(y-b)\; dy=0$ or using the implied expression for $b$: $0=2xy \; dx + (2y^2-2yb) \;dy=2xy\; dx + (y^2-x^2)\; dy$ i.e. the above differential.

Answer 3

@H. H. Rugh an equivalent way to yours that one could find in an old textbook of mine is:

"if a family (F) of curves is characterized as the set of solutions of differential equation $y'=f(x,y)$, the family of orthogonal curves to all curves of (F) is solution of the differential equation

$$-\dfrac{1}{y'}=f(x,y)$$

(of course, in this kind of books, it was implicit that the special case $y'=0$ had to be considered apart...)