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 [math]0<a\le r\in\mathbb{R}[/math].

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:

[math]F(x,y)=C[/math]

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

[math]\frac{\partial F}{\partial y}dx-\frac{\partial F}{\partial x}dy=0[/math]

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:

[math]x^2+(y-k)^2=r^2[/math]

By the Pythagorean theorem, we see that:

[math]k^2=r^2-a^2[/math]

hence:

[math]k=\pm\sqrt{r^2-a^2}[/math]

and so the family of circles is:

[math]x^2+\left(y\mp\sqrt{r^2-a^2} \right)^2=r^2[/math]

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

[math]F(x,y)=k[/math]

[math]x^2+y^2\mp2\sqrt{r^2-a^2}y+r^2-a^2=r^2[/math]

[math]x^2+y^2-a^2=\pm2\sqrt{r^2-a^2}y[/math]

[math]\frac{x^2+y^2-a^2}{y}=\pm2\sqrt{r^2-a^2}[/math]

It can be shown that the orthogonal trajectories must satisfy:

[math]\frac{\partial F}{\partial y}dx-\frac{\partial F}{\partial x}dy=0[/math]

So, we compute:

[math]\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}[/math]

[math]\frac{\partial F}{\partial x}=\frac{2x}{y}[/math]

Hence, the ODE we need to solve is:

[math]\frac{y^2-x^2+a^2}{y^2}dx-\frac{2x}{y}dy=0[/math]

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

[math](y^2-x^2+a^2)\,dx-(2xy)\,dy=0[/math]

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

[math]M(x,y)=y^2-x^2+a^2\,\therefore\,\frac{\partial M}{\partial y}=2y[/math]

[math]N(x,y)=-2xy\,\therefore\,\frac{\partial N}{\partial x}=-2y[/math]

and so:

[math]\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N(x,y)}=\frac{2y-(-2y)}{-2xy}=\frac{4y}{-2xy}=-\frac{2}{x}[/math]

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

[math]\mu(x)=e^{-2\int\frac{dx}{x}}=e^{\ln(x^{-2})}=x^{-2}[/math]

Multiplying the ODE by this integrating factor, we obtain:

[math]\frac{y^2-x^2+a^2}{x^2}dx-\frac{2y}{x}dy=0[/math]

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

[math]F(x,y)=\int\frac{y^2-x^2+a^2}{x^2}\,dx+g(y)[/math]

[math]F(x,y)=-\left(\frac{y^2+a^2}{x}+x \right)+g(y)[/math]

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

[math]-\frac{2y}{x}=-\frac{2y}{x}+g'(y)[/math]

[math]g'(y)=0[/math]

[math]g(y)=C[/math]

And so we have:

[math]F(x,y)=-\left(\frac{y^2+a^2}{x}+x \right)+C[/math]

Hence, the orthogonal trajectories are given by:

[math]\frac{y^2+a^2}{x}+x=C[/math]

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

[math](x-c)^2+y^2=c^2-a^2[/math] where [math]a<|c|[/math]