# Tangent to Astroid between Coordinate Axes has Constant Length

## Theorem

Let $C_1$ be a circle of radius $b$ roll without slipping around the inside of a circle $C_2$ of radius $a = 4 b$.

Let $C_2$ be embedded in a cartesian coordinate plane with its center $O$ located at the origin.

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \left({a, 0}\right)$ on the $x$-axis.

Let $H$ be the astroid formed by the locus of $P$.

The segment of the tangent to $H$ between the $x$-axis and the $y$-axis is constant, immaterial of the point of tangency.

## Proof

From Equation of Astroid, $H$ can be expressed as:

$\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$

Thus the slope of the tangent to $H$ at $\left({x, y}\right)$ is:

 $\displaystyle \frac {\mathrm d y} {\mathrm d x}$ $=$ $\displaystyle \frac {3 a \sin^2 \theta \cos \theta \, \mathrm d \theta} {-3 a \cos^2 \theta \sin \theta \, \mathrm d \theta}$ $\displaystyle$ $=$ $\displaystyle - \tan \theta$

Thus the equation of the tangent to $H$ is given by:

$y - a \sin^3 \theta = -\tan \theta \left({x - a \cos^3 \theta}\right)$

The $x$-intercept is found by setting $y = 0$ and solving for $x$:

 $\displaystyle x$ $=$ $\displaystyle a \cos^3 \theta + a \sin^2 \theta \cos \theta$ $\displaystyle$ $=$ $\displaystyle a \cos \theta \left({\cos^2 \theta + \sin^2 \theta}\right)$ $\displaystyle$ $=$ $\displaystyle a \cos \theta$ Sum of Squares of Sine and Cosine

Similarly, the $y$-intercept is found by setting $x = 0$ and solving for $y$, which gives:

$y = a \sin \theta$

The length of the part of the tangent to $H$ between the $x$-axis and the $y$-axis is given by:

 $\displaystyle \sqrt {a^2 \cos^2 \theta + a^2 \sin^2 \theta}$ $=$ $\displaystyle a \sqrt {\cos^2 \theta + \sin^2 \theta}$ $\displaystyle$ $=$ $\displaystyle a$ Sum of Squares of Sine and Cosine

which is constant.

$\blacksquare$