Analytic Geometry: Orthogonal Trajectories

The problem is as follows:

a) Find the family of circles centered on the $y$-axis, that pass through the points $(\pm a,0)$, where $\displaystyle 0<a\le r\in\mathbb{R}$.

b) Find the family of curves orthogonal to the family of circles found in part a).

Hint 1: Express the family of circles from part a) in the form:

$\displaystyle F(x,y)=C$

Hint 2: It can be shown that the orthogonal trajectories must satisfy:

$\displaystyle \frac{\partial F}{\partial y}dx-\frac{\partial F}{\partial x}dy=0$

My solution:

a) To find the required family of circles, consider that we have the center of the circle at $(0,k)$ where the radius is $r$. Thus the equation of the circle is:

$\displaystyle x^2+(y-k)^2=r^2$

By the Pythagorean theorem, we see that:

$\displaystyle k^2=r^2-a^2$

hence:

$\displaystyle k=\pm\sqrt{r^2-a^2}$

and so the family of circles is:

$\displaystyle x^2+\left(y\mp\sqrt{r^2-a^2} \right)^2=r^2$

b) Now to find the orthogonal trajectories, we will express the family of circles in the form:

$\displaystyle F(x,y)=k$

$\displaystyle x^2+y^2\mp2\sqrt{r^2-a^2}y+r^2-a^2=r^2$

$\displaystyle x^2+y^2-a^2=\pm2\sqrt{r^2-a^2}y$

$\displaystyle \frac{x^2+y^2-a^2}{y}=\pm2\sqrt{r^2-a^2}$

It can be shown that the orthogonal trajectories must satisfy:

$\displaystyle \frac{\partial F}{\partial y}dx-\frac{\partial F}{\partial x}dy=0$

So, we compute:

$\displaystyle \frac{\partial F}{\partial y}=\frac{y(2y)-(x^2+y^2-a^2)(1)}{y^2}=\frac{y^2-x^2+a^2}{y^2}$

$\displaystyle \frac{\partial F}{\partial x}=\frac{2x}{y}$

Hence, the ODE we need to solve is:

$\displaystyle \frac{y^2-x^2+a^2}{y^2}dx-\frac{2x}{y}dy=0$

Multiplying through by $y^2$, we obtain the equation:

$\displaystyle (y^2-x^2+a^2)\,dx-(2xy)\,dy=0$

This equation is not separable, exact nor linear, so let's look at finding an integrating factor that will make it exact. We have:

$\displaystyle M(x,y)=y^2-x^2+a^2\,\therefore\,\frac{\partial M}{\partial y}=2y$

$\displaystyle N(x,y)=-2xy\,\therefore\,\frac{\partial N}{\partial x}=-2y$

and so:

$\displaystyle \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N(x,y)}=\frac{2y-(-2y)}{-2xy}=\frac{4y}{-2xy}=-\frac{2}{x}$

Since this is a function of $x$ alone, our integrating factor is:

$\displaystyle \mu(x)=e^{-2\int\frac{dx}{x}}=e^{\ln(x^{-2})}=x^{-2}$

Multiplying the ODE by this integrating factor, we obtain:

$\displaystyle \frac{y^2-x^2+a^2}{x^2}dx-\frac{2y}{x}dy=0$

Now we have an exact equation. If the solution is $F(x,y)=C$, then we must have:

$\displaystyle F(x,y)=\int\frac{y^2-x^2+a^2}{x^2}\,dx+g(y)$

$\displaystyle F(x,y)=-\left(\frac{y^2+a^2}{x}+x \right)+g(y)$

Next, to determine $g(y)$, we will take the partial derivative with respect to $y$ and substitute $\displaystyle \frac{\partial F}{\partial y}=-\frac{2y}{x}$ and solve for $g'(y)$.

$\displaystyle -\frac{2y}{x}=-\frac{2y}{x}+g'(y)$

$\displaystyle g'(y)=0$

$\displaystyle g(y)=C$

And so we have:

$\displaystyle F(x,y)=-\left(\frac{y^2+a^2}{x}+x \right)+C$

Hence, the orthogonal trajectories are given by:

$\displaystyle \frac{y^2+a^2}{x}+x=C$

Multiplying through by $x$ and completing the square, and using $\displaystyle c=\frac{C}{2}$, we find the orthogonal trajectories is given by the family of circles:

$\displaystyle (x-c)^2+y^2=c^2-a^2$ where $\displaystyle a<|c|$