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Why does the following equation draw a circle ?

$$\left(\frac{t^4-6t^2+1}{t^4+2t^2+1},\frac{4t-4t^3}{t^4+2t^2+1}\right),|t|\le1$$

Does it draw a perfect circle, or an approximation ? On Desmos, it looks like a perfect circle.

(Added by edit) If it is an exact equation, how does one find such an equation ? Where does it come from ?

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  • $\begingroup$ What properties does a perfect circle have? $\endgroup$ Commented Jan 10, 2015 at 19:00
  • $\begingroup$ All the points are equidistant from $0$ @Alizter $\endgroup$ Commented Jan 10, 2015 at 19:01
  • $\begingroup$ Also think about the tangent at a point $\endgroup$ Commented Jan 10, 2015 at 19:01
  • $\begingroup$ @Alizter $\sqrt{r^2-x^2}'=\dfrac{-x}{\sqrt{r^2-x^2}}$ $\endgroup$ Commented Jan 10, 2015 at 19:02
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    $\begingroup$ So find $dx/dt$ and $dy/dt$ and take the 'ratio' $(dy/dt)/(dx/dt)$ to equate to your gradient $\endgroup$ Commented Jan 10, 2015 at 19:04

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The rational parametrization of the unit circle that’s most often seen is $$ x=\frac{2t}{t^2+1}\,,\quad y=\frac{t^2-1}{t^2+1}\,. $$ You can get this, as I recall, by drawing the line through $(0,-1)$ with slope $t$ and seeing where it intersects the unit circle. It’s a nice exercise.

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    $\begingroup$ With the given mode of generation, a sign change is needed for $y$ : $y=(1-t^2)/(1+t^2)$ $\endgroup$
    – Jean Marie
    Commented Jan 11 at 10:54
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Hint: check that the quantity $x(t)^2 + y(t)^2$ is constant. This proves that the image of your curve is contained in a circumference of radius $r = \sqrt{ x(t)^2 + y(t)^2}$ centered at the origin.

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As the previous posters noticed, it is easy to check that $x(t)^2+y(t)^2=1.$ That is if we choose a point $(a,b)$ of the unit circle there is some $t$ such that $x(t)=a,y(t)=b.$

Assume that all the candidates $t$ has $|t|>1.$ Then the same holds for $$(t^4-6t^2+1)=a(t^2+1)^2,-4t(t^2-1)=b(t^2+1)^2.$$ If we eliminate $t^2+1 $(assume $b\not=0)$ then the equation $f(t)=(t^4-6t^2+1)b+4t(t^2-1)a=0$ is satisfied only for $t>1.$ But this leads to a contradiction since $f(t)=0$ has a solution in $[0,1]$ (and [-1,0]) since $f(0)f(1)=-4b^2<0.$ So always we can find $t$ with $|t|\leq 1$ such that, given $(a,b)$ on the circle we have $x(t)=a,y(t)=b.$ Now for the case $b=0$ we can choose $t=0$ if $a=1$ and $t=1$ if $a=-1$.

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