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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.

Gennaro Arguzzi
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  • I think you could just convert to cartesian, where it will be obvious. – John Lou Jun 24 '17 at 17:15
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    You need some information about the parametric equation to prove that is describes a circle (or even part of a circle). A typical approach would be to find a center $(h,k)$ and prove that $(x-h)^2 + (y-k)^2$ is a positive constant (radius squared) as time $t$ varies. – hardmath Jun 24 '17 at 17:16

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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.

paw88789
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  • Hi @paw88789. For example, how can I prove that, when t chenges, $\left(\left(cos(t)\right),\left(sin(t)\right)\right)$ describes the whole circle and not only a part? – Gennaro Arguzzi Jun 25 '17 at 07:56
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    @GennaroArguzzi For instance if $t$ goes from $0$ to $2\pi$, then you will start and end at $(1,0)$, corresponding to one full traversal of the circle. (By checking the values of $x(t)$ and $y(t)$ at the ends of the parameter interval.) – paw88789 Jun 25 '17 at 13:32
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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.

Chappers
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  • Hi @Chappers. I have two questions for you. 1. Where comes from the term $\frac{\sqrt{x'^2+y'^2}}{\lvert x' \rvert}$ at the beginning of proof? 2. What is the difference between the derivatives $x'$ and $\dot{x}$? Thank you. – Gennaro Arguzzi Jun 25 '17 at 06:30
  • Square both sides of the left equation, add $1$ and take the square root. Sorry if that wasn't as obvious as I thought! 2. I meant $x'$ to mean $\frac{d}{dt} x(t)$ and $\dot{x}$ to be $\frac{d}{ds} x(t(s)) $, but I suppose it's not that clear. I'll edit.
    • – Chappers Jun 25 '17 at 11:13
  • Very nice and complete solution (+1). I am intrigued to see the appearance of the rarely-encountered $\cdot^{3/2}$ (power) which also appears in the solution here on the length of the latus rectum of a parametric parabola. Could there be any relationship between the two? – Hypergeometricx Jun 25 '17 at 16:32
  • @hypergeometric I don't think they're related: the parabola expression is effectively $(x''y'-y''x')^2/(x''^2+y''^2)^{3/2}$ (which, in particular, is constant in that case). – Chappers Jun 25 '17 at 21:04
  • @hypergeometric And rotating and reparametrising the original expression into the standard form $(x_0,y_0)+(\alpha t^2,2\alpha t)$ gives $\alpha=(aB-Ab)^2/(a^2+A^2)^{3/2}$ anyway. – Chappers Jun 25 '17 at 21:19
  • OK. Just that one doesn't often come across $(x^2+y^2)^{3/2}$. – Hypergeometricx Jun 26 '17 at 00:33