Epicycloid and Hypocycloid

Summary: §§523–524. Curves traced by a point on (or extending from) a circle that rolls around the outside (epicycloid) or inside (hypocycloid) of a stationary circle. Parametric equations $z=(a+c)\cos (s/c)+b\cos ((a+c)s/(ac))$ and $y$ analogue with sines. Algebraic when $(a+c)/a$ is rational; transcendental otherwise. Special case $b=-(a+c)$ collapses to a curve through the center; $c=2a$ yields the sixth-degree curve $(z^2+y^2{)}^3-2f^2(z^2+y^2{)}^2+f^4{z}^2=0$. Drawn as figure 106.

Sources: chapter21 §§523–524, figures106-108 (figure 106).

Last updated: 2026-05-12

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Construction (§523, figure 106)

Setup:

• Stationary circle of radius $OA=c$, centered at $O$.
• Moving circle of radius $a$ rolling on the outside, with center $C$ at distance $OC=a+c$ from $O$.
• Tracing point $D$ on the extended diameter of the moving circle at distance $CD=b$.

The desired curve $Dd$ uses $OD$ as axis. From the initial position (where $O,C,D$ are collinear), the moving circle rolls so its contact point is at $Q$ on the stationary circle, with arc $AQ=z$ subtending angle $z/c$ at $O$. The moving circle is now in position $QcR$, and since arcs rolled equal:

$\text{arc }Qa=z\Rightarrow \angle acQ=z/a,\angle Rcd=z/a,cd=b.$

The angle $dcn$ (where $cn$ is parallel to $OD$) is therefore $z/c+z/a=(a+c)z/(ac)$.

Drop perpendiculars and combine:

$dn=b\sin \frac{(a+c)z}{ac},cn=b\cos \frac{(a+c)z}{ac}.$

The center $C$ of the moving circle moves with $Oc=a+c$, so

$cm=(a+c)\sin (z/c),Om=(a+c)\cos (z/c).$

Setting $OP=z^{\prime }$ (the new abscissa, distinct from the arc $z$) and $Pd=y$ and dropping the prime:

$\boxed{z=(a+c)\cos \frac{z^{\prime }}{c}+b\cos \frac{(a+c)z^{\prime }}{ac},y=(a+c)\sin \frac{z^{\prime }}{c}+b\sin \frac{(a+c)z^{\prime }}{ac}.}$

(Where $z^{\prime }$ here parametrizes by arc, and I am breaking Euler’s overloaded notation: he uses $z$ both for arc and abscissa.)

When is it algebraic? (§523)

It should be clear that if $(a+c)/a$ is a rational number, then because of the commensurability of the angle $z/c$ and $(a+c)z/(ac)$, the unknown quantity $z$ can be eliminated so that it is possible to find an algebraic equation in $z$ and $y$. In the other cases the curve described in this way will be transcendental. (source: chapter21, §523)

This is the central observation: rational ratio $\Rightarrow$ algebraic curve (the moving circle closes up after finitely many rotations); irrational $\Rightarrow$ transcendental (the path never closes).

Examples of the rational case. If $a=c$ (moving and stationary radii equal), $(a+c)/a=2$ — closes after one revolution, gives the cardioid (or epitrochoid for $b\ne a$). If $a=c/3$, gives the astroid family.

Hypocycloid case

Here note that if we take $a$ to be negative, then the result will be a hypocycloid, since the moving circle lies inside the stationary circle. (source: chapter21, §523)

So the same formulas, with $a<0$, describe the hypocycloid. Proper epicycloids/hypocycloids take $b=a$ (point on the circumference); $b\ne a$ gives the epitrochoid/hypotrochoid.

A useful identity (§523)

Squaring and adding:

$z^2+y^2=(a+c{)}^2+b^2+2b(a+c)\cos (z/a).$

This expresses $\cos (z/a)$ in $z^2+y^2$, useful for eliminating $z$ when $a$ and $c$ are commensurable.

Special case $b=-(a+c)$ (§524)

If $b=-(a+c)$, the tracing point $D$ falls on the center $O$ of the stationary circle. The curves pass through $O$. From the identity above:

$z^2+y^2=2(a+c{)}^2\left(1-\cos \frac{z}{a}\right)=4(a+c{)}^2{\cos }^2\frac{z}{2a}.$

Hence $\cos (z/(2a))=\sqrt{z^2+y^2}/(2(a+c))$. From the parametric forms:

$\frac{z}{y}=-\tan \frac{(2a+c)z}{2ac},\sin \frac{(2a+c)z}{2ac}=\frac{z}{\sqrt{z^2+y^2}},\cos \frac{(2a+c)z}{2ac}=\frac{-y}{\sqrt{z^2+y^2}}.$

And $\sqrt{z^2+y^2}=2(a+c)\cos (z/(2a))$, so

$z=2(a+c)\cos \frac{z}{2a}\sin \frac{(2a+c)z}{2ac},y=-2(a+c)\cos \frac{z}{2a}\cos \frac{(2a+c)z}{2ac}.$

Sub-special case $c=2a$: then $\cos (z/(2a))$ and the second factors simplify; let $q=\cos (z/(2a))$, so $\sin (z/(2a))=\sqrt{1-q^2}$, $\sin (z/a)=2q\sqrt{1-q^2}$, $\cos (z/a)=2q^2-1$. Then

$y=6aq(2q^2-1)=(1-2q^2)\sqrt{z^2+y^2},q=\frac{\sqrt{z^2+y^2}}{6a},$

and

$18a^{2}y=(18a^2-z^2-y^2)\sqrt{z^2+y^2}.$

Let $18a^2=f^2$ and square:

$\boxed{(z^2+y^2{)}^3-2f^2(z^2+y^2{)}^2+f^4{z}^2=0.}$

A sextic, hence algebraic for this rational-ratio case (consistent with $(a+c)/a=3$). Euler closes:

Since we are now concerned with transcendental rather than algebraic curves, we will leave this topic and move on to curves whose construction requires both logarithms and circular arcs. (source: chapter21, §524)

Figures

Figures 106–108

Related pages

• chapter-21-on-transcendental-curves
• cycloid — natural specialization (rolling on a line instead of a circle).
• arccos-log-curve — next member of the chapter’s catalogue, blending logs and arcs.
• similar-curves — affinity-and-similarity perspective on the rational/irrational dichotomy.
• transcendental-curves