Parametric equation of a circle - Proof

Question

I know that a circle can be described with infinite parametric equations. Given a parametric equation $(x(t), y(t))$, how can I prove that the curve is a circle?

Thank you very much.

Answer 1

One possibility would be to show that for all $t$, $x(t)^2+y(t)^2$ is the same positive real number. You would also have to show that for an appropriate parameter interval, the entire circle is generated.

Answer 2

From the standard definition of a circle:

you have to prove that there exists a point $C$ that has the same distance from any pointwit the coordinates $(x(t),y(t))$

Answer 3

This answer shows what to do if we don't know that the circle is centred at zero. A curve is part of a circle if and only if its radius of curvature is a finite constant (and hence so is its curvature):

A circle satisfies $(x-a)^2+(y-b)^2=c^2$. Differentiating, $$ (x-a)x' + (y-b)y' = 0, $$ or $$ \frac{x-a}{y-b} = -\frac{y'}{x'} \implies \frac{\sqrt{x'^2+y'^2}}{\lvert x' \rvert} = \frac{\sqrt{(y-b)^2+(x-a)^2}}{\lvert y-b \rvert} = \frac{c}{\lvert y-b \rvert}, $$ (squaring both sides, adding $1$ and taking a square root) and differentiating the first of these again, $$ \frac{-y''x'+x''y'}{x'^2} = \frac{x'}{y-b} -\frac{(x-a)y'}{(y-b)^2} = \frac{1}{y-b} \left( x' + \frac{y'^2}{x'} \right) = \frac{x'^2+y'^2}{(y-b)x'}, $$ so $$ \left| \frac{x''y'-y''x'}{x'^2+y'^2} \right| = \frac{\lvert x' \rvert}{\lvert y-b \rvert} = \frac{\sqrt{x'^2+y'^2}}{c}. $$ Thus $$ \left| \frac{x''y'-y''x'}{(x'^2+y'^2)^{3/2}} \right| = \frac{1}{c}. $$ It therefore is sensible to call the reciprocal of this quantity the radius of curvature.

Now, let's show the other direction: that if $\left| \frac{x''y'-y''x'}{(x'^2+y'^2)^{3/2}} \right|$ is constant, $(x,y)$ describes an arc of a circle. It is useful to know how the curvature changes under reparametrisation: if $t=t(s)$, and we use the convention that $' = d/dt$, $\dot{} = d/ds$, then $$\dot{x}^2+\dot{y}^2 = \dot{t}^2(x'^2+y'^2),$$ and $$ \ddot{x}\dot{y}-\ddot{y}\dot{x} = (x'\ddot{t}+x''\dot{t}^2)y'\dot{t} - (y'\ddot{t}+y''\dot{t}^2)x'\dot{t} = \dot{t}^3(x''y'-y''x'), $$ and so the curvature is invariant under reparametrisation, which is sensible.

It's especially nice because we can re-parametrise using the arc-length, where $x'^2+y'^2=1$, so it amounts to solving $$ x''y'-y''x' = 1/c. $$ Differentiating the arc-length equation gives $$ x''x'+y''y' = 0 \implies y''= -\frac{x''x'}{y'}, $$ so $$ \frac{1}{c} = x''y' +\frac{x''x'^2}{y'} = x''\frac{x'^2+y'^2}{y'} = \frac{x''}{y'}, $$ so $y'=cx''$ and $x'=-cy''$, a system of equations which, together with $x'^2+y'^2=1$, has only solutions of the form $x=a+c\cos{((t-t_0)/c)}$, $y=b+c\sin{((t-t_0)/c)}$, i.e. circular arcs.