Hello Nikachu,
To find the required family of circles, please refer to the following diagram:
We have the center of the circle at $(0,k)$ where the radius is $r$. Thus the equation of the circle is:
$$x^2+(y-k)^2=r^2$$
By the Pythagorean theorem, we see that:
$$k^2=r^2-1$$
hence:
$$k=\pm\sqrt{r^2-1}$$
and so the family of circles is:
$$x^2+\left(y\mp\sqrt{r^2-1} \right)^2=r^2$$
Now to find the orthogonal trajectories, we will express the family of circles in the form:
$$F(x,y)=k$$
$$x^2+y^2\mp2\sqrt{r^2-1}y+r^2-1=r^2$$
$$x^2+y^2-1=\pm2\sqrt{r^2-1}y$$
$$\frac{x^2+y^2-1}{y}=\pm2\sqrt{r^2-1}$$
It can be shown that the orthogonal trajectories must satisfy:
$$\frac{\delta F}{\delta y}dx-\frac{\delta F}{\delta x}dy=0$$
So, we compute:
$$\frac{\delta F}{\delta y}=\frac{y(2y)-(x^2+y^2-1)(1)}{y^2}=\frac{y^2-x^2+1}{y^2}$$
$$\frac{\delta F}{\delta x}=\frac{2x}{y}$$
Hence, the ODE we need to solve is:
$$\frac{y^2-x^2+1}{y^2}dx-\frac{2x}{y}dy=0$$
Multiplying through by $y^2$, we obtain the equation:
$$(y^2-x^2+1)\,dx-(2xy)\,dy=0$$
This equation is not separable, exact nor linear, so let's look at finding an integrating factor that will make it exact. We have:
$$M(x,y)=y^2-x^2+1\,\therefore\,\frac{\delta M}{\delta y}=2y$$
$$N(x,y)=-2xy\,\therefore\,\frac{\delta N}{\delta x}=-2y$$
and so:
$$\frac{\frac{\delta M}{\delta y}-\frac{\delta N}{\delta x}}{N(x,y)}=\frac{2y-(-2y)}{-2xy}=\frac{4y}{-2xy}=-\frac{2}{x}$$
Since this is a function of $x$ alone, our integrating factor is:
$$\mu(x)=e^{-2\int\frac{dx}{x}}=e^{\ln(x^{-2})}=x^{-2}$$
Multiplying the ODE by this integrating factor, we obtain:
$$\frac{y^2-x^2+1}{x^2}dx-\frac{2y}{x}dy=0$$
Now we have an exact equation. If the solution is $F(x,y)=C$, then we must have:
$$F(x,y)=\int\frac{y^2-x^2+1}{x^2}\,dx+g(y)$$
$$F(x,y)=-\left(\frac{y^2+1}{x}+x \right)+g(y)$$
Next, to determine $g(y)$, we will take the partial derivative with respect to $y$ and substitute $$\frac{\delta F}{\delta y}=-\frac{2y}{x}$$ and solve for $g'(y)$.
$$-\frac{2y}{x}=-\frac{2y}{x}+g'(y)$$
$$g'(y)=0$$
$$g(y)=C$$
And so we have:
$$F(x,y)=-\left(\frac{y^2+1}{x}+x \right)+C$$
Hence, the orthogonal trajectories are given by:
$$\frac{y^2+1}{x}+x=C$$
Multiplying through by $x$ and completing the square, and using $$c=\frac{C}{2}$$, we find the orthogonal trajectories is given by the family of circles:
$$(x-c)^2+y^2=c^2-1$$ where $$1<|c|$$
To Nikachu and any other guests viewing this topic, I invite and encourage you to post other differential equations problems here in our http://www.mathhelpboards.com/f17/ forum.
Best Regards,
Mark.