Circles, sectors, and ellipses: area and circumference

The elementary table lists a circle, a sector of a circle, and an ellipse, each with its own $A$ and $C$. Three names look like three topics. The sector is the circle with the central angle frozen below a full turn. The ellipse is the circle stretched along two axes. $C$ is the circumference. This math handbook writes the central angle $\alpha$ in radians; a full turn is $2\pi$. See measuring angles.

One radius

A radius $r$ from the centre to a point on the circle determines both measures:

$A=\pi r^2,C=2\pi r.$

No extra freeze: the full turn is already $2\pi$. Drag $A$; the segment $OA$ is $r$.

Cut a wedge

Keep the same circle and freeze a central angle $\alpha$ between two radii. Let $L$ be the arc between them:

$A=\frac{1}{2}\alpha r^2,L=\alpha r,C=L+2r.$

The extra freeze is the wedge: the boundary $C$ walks the arc and both radii. When $\alpha =2\pi$, $A$ and $L$ become the circle's; the extra $+2r$ is those two radii, which now coincide, so $C$ is not the circle formula. Drag $B$ on the circle to change $\alpha$. Drag $A$ to change $r$; $B$ stays on the circle.

Stretch the two axes

Keep the disk and freeze two different semi-axes $a$ and $b$, with $b<a$. Area fills vertical strips: flatten every height by $b/a$ and the whole disk scales by that one factor,

$A=\pi ab.$

The same stretch splits the centre into two foci on the major axis. Their distance from the centre is $\epsilon a$, where $\epsilon$ is the numerical eccentricity

$\epsilon =\sqrt{1-\frac{b^2}{a^2}}.$

On a circle the foci sit on the centre, so $\epsilon =0$. That is the extra freeze $a=b$ dropped. Drag $P$ toward the major axis: $A$ stays put, so $a$ is fixed, $b$ shrinks, and the foci run outward. The dashed segment is $\epsilon a$. Keep $P$ closer to $O$ than $A$ is, or $b<a$ fails.

Length does not stretch the same way

The stretch that gave $A=\pi ab$ does not give the rim. Length cares about direction. A bit of the circle that ran along the $x$-axis keeps its length; a bit that ran up the $y$-axis shrinks; a diagonal bit mixes the two. There is no single multiple of $2\pi a$ that is $C$.

So you add the changed bits as you go around — an integral. That integral cannot be written from $a$ and $b$ by finitely many arithmetic operations and roots; it is named $E(\epsilon )$.

Four equal quarters make the rim. Measure the quarter from the end of the major axis to the end of the minor axis in units of $a$:

$\text{one quarter}=aE(\epsilon ),C=4aE(\epsilon ).$

$E(\epsilon )$ is not a length — it is how many times $a$ fits into that quarter. On a circle the quarter is a quarter-circle, so $E(0)=\pi /2$ and $C=2\pi a$. Flatten with $a$ fixed and the quarter shrinks toward $a$, so $E$ falls toward $1$. Every ellipse has $0\le \epsilon <1$.

The outer circle has radius $a$; the inner dashed circle has radius $b$; the ellipse sits between them. From $Q$ on the outer circle, drop a vertical to the ellipse at $R$. Then $R=(a\cos \phi ,b\sin \phi )$, where $\phi =\angle AOQ$ is the eccentric angle — the polar angle of $Q$, not of $R$. That extra naming is why this board is not the same as dragging a point on the ellipse.

A step $d\phi$ moves $Q$ through an arc $ad\phi$. The same step moves $R$ through

$ds=a\sqrt{1-{\epsilon }^2{\cos }^2\phi }d\phi .$

At $A$ ($\phi =0$) that speed is $b$; at the top ($\phi =\pi /2$) it is $a$. Integrate from $A$ to the top and divide by $a$. Replace $\phi$ by $\pi /2-\phi$ and $\cos$ becomes $\sin$; the definite integral does not change, and that number is the complete elliptic integral of the second kind

$E(\epsilon ):={\int }_0^{\pi /2}\sqrt{1-{\epsilon }^2{\sin }^2\phi }d\phi .$

The limits $0$ to $\pi /2$ are what complete means: the full quarter, not a stop at a smaller $\phi$. Second kind is this integrand; the first kind puts the square root in the denominator and is not an arc length.

Drag $Q$ in the first quadrant. $R$ traces the quarter; four of those quarters make $C$. Drag $B$ toward $O$ to flatten: $\epsilon$ grows, $R$ is slower near $A$, and $E$ falls below $\pi /2$. The ellipse stays between the two circles, and $2\pi b<C<2\pi a$.

When $\epsilon =0$, $B$ reaches the outer circle, the drop has length zero, $R$ stays on $Q$, and you integrate $1$ — so $E=\pi /2$ and $C=2\pi a$. Near a circle the first correction is quadratic:

$E(\epsilon )=\frac{\pi }{2}\left(1-\frac{{\epsilon }^2}{4}-\frac{3{\epsilon }^4}{64}-\frac{5{\epsilon }^6}{256}-\cdots \right),$

hence $C\approx 2\pi a(1-{\epsilon }^2/4)$. Flatten a little and ${\epsilon }^2/4$ is the relative drop to expect before the higher terms matter.

Polygons from the same table live under triangles, quadrilaterals, and regular n-gons.