# Orthogonal Trajectories/Circles Tangent to Y Axis/Proof 2

## 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

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.

$\blacksquare$