Orthogonal Trajectories/Examples/Circles Tangent to Y Axis

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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.


CirclesTangentAxisOrthogonalTrajectories.png


Proof 1

We use the technique of formation of ordinary differential equation by elimination.

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

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

from which:

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

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

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


Let:

$\map M {x, y} = 2 x y$
$\map N {x, y} = x^2 - y^2$


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

\(\ds \map M {t x, t y}\) \(=\) \(\ds 2 t x t y\)
\(\ds \) \(=\) \(\ds t^2 \paren {2 x y}\)
\(\ds \) \(=\) \(\ds t^2 \, \map M {x, y}\)
\(\ds \map N {t x, t y}\) \(=\) \(\ds \paren {t x}^2 - \paren {t y}^2\)
\(\ds \) \(=\) \(\ds t^2 \map N {x^2 - y^2}\)
\(\ds \) \(=\) \(\ds t \map N {x, y}\)

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:

$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:

$\map f {x, y} = \dfrac {2 x y} {x^2 - y^2}$

Thus:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {2 z} {1 - z^2} - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {1 - z^2} {z \paren {1 + z^2} } \rd z + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {\d z} {z \paren {1 + z^2} } \rd z - \int \frac z {\paren {1 + z^2} } \rd z + C_1\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {\frac {z^2} {z^2 + 1} } - \frac 1 2 \map \ln {z^2 + 1} + C_1\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {\frac {z^2} {\paren {z^2 + 1}^2} } + C_1\)
\(\ds \leadsto \ \ \) \(\ds C_2 x^2\) \(=\) \(\ds \frac {z^2} {\paren {z^2 + 1}^2}\)
\(\ds \leadsto \ \ \) \(\ds C_3 x\) \(=\) \(\ds \frac {y / x} {\paren {y / x}^2 + 1}\)
\(\ds \leadsto \ \ \) \(\ds x^2 + y^2\) \(=\) \(\ds 2 C y\)

$\blacksquare$


Proof 2

We use the technique of formation of ordinary differential equation by elimination.

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 Solution to Separable Differential Equation:

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

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.

$\blacksquare$


Sources

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