Orthogonal trajectories of circle Find the orthogonal trajectories of follow circle.
$$(x+C)^2+y^2=C^2$$
My step is follow :

*

*Find y' of original circle : $x+C+yy'=0,\ C={x^2+y^2\over -2x},\ y'={y^2-x^2\over 2xy}$

*Replace y' to $-1/y'$ : $y'={2xy\over x^2-y^2}$
next is problem. How can I solve the equation $y'={2xy\over x^2-y^2}$ ?

Alternatively, I will show you how to solve this using polar coordinates (see section In polar coordinates, wiki).
$x = r \cos\theta, y = r \sin \theta$
$(x+C)^2 + y^2 = C^2 \implies r = - 2 C \cos\theta$
[For $C \gt 0, \pi/2 \leq \theta \leq 3 \pi/2$ and for $C \lt 0, -\pi/2 \leq \theta \leq \pi/2$]
For $C \ne 0$, these are circles centered on x-axis with radius equal to the distance of the center from the origin i.e. y-axis is tangent to these circles.
As the pencil of curves is represented in polar coordinates by
$F(r, \theta, C) = r + 2 C \cos\theta = 0$
$F_r(r, \theta, C) + F_{\theta} (r, \theta, C) \cdot \theta' = 1 - 2C \sin \theta \cdot d\theta' = 0$
Substituting $C = - \cfrac{r}{2 \cos\theta}$ from $(1)$, we get
$1 + r \tan \theta \cdot d\theta = 0 \implies \theta' = f(r, \theta) = - \cfrac{\cot\theta}{r}$
So the differential equation of the orthogonal trajectories is,
$\theta' = - \cfrac{1}{r^2 f(r, \theta)} = \cfrac{1}{r \cot\theta}$
Integrating, $\ln r = \ln (\sin\theta) + c_1$
$\ln r = \ln (2 d \sin \theta) \implies r = 2 d  \sin\theta$
For $d \gt 0, 0 \leq \theta \leq \pi$ and for $d \lt 0, \pi \leq \theta \leq 2\pi$
In cartesian coordinates, you can write the equation of the circle as $x^2 + (y - d)^2 = d^2$
For $d \ne 0$, these are circles centered on y-axis with radius equal to the distance of the center from the origin i.e. x-axis is tangent to the circles. You can see why they are orthogonal trajectories to $(x+C)^2 + y^2 = C^2$.
