Circles in the plane

An ellipse needs two foci and a constant sum $2a$ before the oval closes. A hyperbola freezes a difference and splits; a parabola freezes one focus against a line. The question is which freeze is even thinner — one point, one length — and still a closed curve, and which extra placement then lets us write a Cartesian equation at all.

That length is the same two-point distance as on lines in the plane. Area and circumference sit with circles, sectors, and ellipses.

One center, one radius

Fix a point $C=(c,d)$ and a length $r>0$. A point $P=(x,y)$ lies on this circle when the distance $CP$ is constantly $r$. There is no second focus on this page. Drag $C$ or $P$ on the first board. The segment is that radius; move either end and the circle is the set of points at the new length, not a new kind of curve.

Two foci and a string of length $2a$ are the ellipse page. This page keeps one center and one $r$.

Write the distance in coordinates

The freeze $CP=r$ does not yet look like an algebraic equation. Square both sides (the distance formula from the lines page) and the same set of $(x,y)$ satisfies

$(x-c{)}^2+(y-d{)}^2=r^2.$

That is not a second definition. It is the radius, written so a Cartesian plane can check it. Leave $C$ off the origin, or change $r$, and the left-hand side still has to carry $c$, $d$, and $r$; dropping them is a further freeze, next.

Sit at the origin, radius one

Extra assumptions: the center sits at the origin and $r=1$. Then the same equation collapses to

$x^2+y^2=1.$

Every other circle on this page is a translate and a scale of this one. Drag $P$. The center does not move; the radius stays $1$. Drop either freeze — move the center, or take $r\ne 1$ — and $x^2+y^2=1$ is the wrong formula for the same geometric circle.

The tangent is that equation, linearized

At a point $(x_0,y_0)$ already on $(x-c{)}^2+(y-d{)}^2=r^2$, the tangent line is

$(x-c)(x_0-c)+(y-d)(y_0-d)=r^2.$

This is not a new curve. It is the Cartesian equation made linear at that point. On the first board that line through $P$ is also perpendicular to the radius $CP$. The right angle is the geometric debt of the algebraic freeze: the radius is the unique direction in which the distance from $C$ is changing, so the tangent must be the directions in which it is not. Leave $P$, and the same $C$ and $r$ still name the same circle; only the line changes.

One $t$ walks the whole curve

The circle is closed, so one parameter can walk all of it:

$x=c+r\cos t,y=d+r\sin t,0\le t<2\pi .$

This math handbook measures $t$ in radians. Read $t$ as time. Then $t=0$ is the point $P$ on the first board — to the right of the center, at $(c+r,d)$. As $t$ runs from $0$ to $2\pi$, $P$ travels counterclockwise (the positive orientation in mathematics) once around the circle.

An ellipse uses the same $\cos t$ and $\sin t$ with two lengths $a$ and $b$. Here there is only $r$. Drop $0\le t<2\pi$ and you have not finished the loop.

Curvature is constant

For a circle of radius $R$, the curvature is, by definition,

$K=\frac{1}{R}.$

That is the remaining bill of “one radius”: how fast the curve turns does not depend on which point you stand at. On an ellipse, a hyperbola, or a parabola the inscribed circle at a point can change; those radii live on polar coordinates and conic sections. On this page $R$ is the same $r$ you already dragged.

The object is the one-center freeze: distance $r$ from $C$. In coordinates that freeze is $(x-c{)}^2+(y-d{)}^2=r^2$. Sit at the origin with $r=1$ and it is the unit circle. The tangent (radius perpendicular), the parameterization, and $K=1/R$ are that equation at a point, along the loop, and as a constant turning.

Stretch one axis ($a\ne b$) and the same closed-curve language becomes the ellipse.