# Orthogonal Trajectories/Circles Tangent to Y Axis

## Theorem

Consider the one-parameter family of curves:

$(1): \quad x^2 + y^2 = 2 c x$

which describes the loci of circles tangent to the $y$-axis at the origin.

Its family of orthogonal trajectories is given by the equation:

$x^2 + y^2 = 2 c y$

which describes the loci of circles tangent to the $x$-axis at the origin. ## Proof 1

Differentiating $(1)$ with respect to $x$ gives:

$2 x + 2 y \dfrac {\mathrm d y} {\mathrm d x} = 2 c$

from which:

$\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {y^2 - x^2} {2 x y}$

Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:

$\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {2 x y} {x^2 - y^2}$

Let:

$M \left({x, y}\right) = 2 x y$
$N \left({x, y}\right) = x^2 - y^2$

Put $t x, t y$ for $x, y$:

 $\displaystyle M \left({t x, t y}\right)$ $=$ $\displaystyle 2 t x t y$ $\displaystyle$ $=$ $\displaystyle t^2 \left({2 x y}\right)$ $\displaystyle$ $=$ $\displaystyle t^2 M \left({x, y}\right)$
 $\displaystyle N \left({t x, t y}\right)$ $=$ $\displaystyle \left({t x}\right)^2 - \left({t y}\right)^2$ $\displaystyle$ $=$ $\displaystyle t^2 N \left({x^2 - y^2}\right)$ $\displaystyle$ $=$ $\displaystyle t N \left({x, y}\right)$

Thus both $M$ and $N$ are homogeneous functions of degree $2$.

Thus, by definition, $(1)$ is a homogeneous differential equation.

By Solution to Homogeneous Differential Equation, its solution is:

$\displaystyle \ln x = \int \frac {\mathrm d z} {f \left({1, z}\right) - z} + C$

where:

$f \left({x, y}\right) = \dfrac {2 x y} {x^2 - y^2}$

Thus:

 $\displaystyle \ln x$ $=$ $\displaystyle \int \frac {\mathrm d z} {\dfrac {2 z} {1 - z^2} - z} + C_1$ $\displaystyle$ $=$ $\displaystyle \int \frac {1 - z^2} {z \left({1 + z^2}\right)} \, \mathrm d z + C_1$ $\displaystyle$ $=$ $\displaystyle \int \frac {\mathrm d z} {z \left({1 + z^2}\right)} \, \mathrm d z - \int \frac z {\left({1 + z^2}\right)} \, \mathrm d z + C_1$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 \ln \left({\frac {z^2} {z^2 + 1} }\right) - \frac 1 2 \ln \left({z^2 + 1}\right) + C_1$ $\displaystyle$ $=$ $\displaystyle \frac 1 2 \ln \left({\frac {z^2} {\left({z^2 + 1}\right)^2} }\right) + C_1$ $\displaystyle \implies \ \$ $\displaystyle C_2 x^2$ $=$ $\displaystyle \frac {z^2} {\left({z^2 + 1}\right)^2}$ $\displaystyle \implies \ \$ $\displaystyle C_3 x$ $=$ $\displaystyle \frac {y/x} {\left({y/x}\right)^2 + 1}$ $\displaystyle \implies \ \$ $\displaystyle x^2 + y^2$ $=$ $\displaystyle 2 C y$

$\blacksquare$

## Proof 2

Expressing $(1)$ in polar coordinates, we have:

$(2): \quad r = 2 c \cos \theta$

Differentiating $(1)$ with respect to $\theta$ gives:

$(3): \quad \dfrac {\d r} {\d \theta} = -2 c \sin \theta$

Eliminating $c$ from $(2)$ and $(3)$:

$r \dfrac {\d \theta} {\d r} = -\dfrac {\cos \theta} {\sin \theta}$

Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:

$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {\cos \theta}$

Using the technique of Separation of Variables:

$\displaystyle \int \frac {\d r} r = \int \dfrac {\cos \theta} {\sin \theta} \rd \theta$

which by Primitive of Reciprocal and various others gives:

$\ln r = \map \ln {\sin \theta} + \ln 2 c$

or:

$r = 2 c \sin \theta$

This can be expressed in Cartesian coordinates as:

$x^2 + y^2 = 2 c y$

Hence the result.

$\blacksquare$