Converting circle parametric equation

Question

I have this parametric equations: $$ x=\frac{a^2-2}{a^2+1} $$ $$ y=\frac{-3a}{a^2+1} $$ for $a\in(-\infty,\infty)$ They describe a circle with center at $(-\frac{1}{2},0)$ and radius $\frac{3}{2}$. Is there any way to convert them into one cartesian equation?

Answer 1

WLOG $a=\tan t$

$\implies y=-\dfrac{3\sin2t}2$ or $\dfrac{2a}{1+a^2}=-\dfrac{2y}3$

and $x=\dfrac{a^2-2}{a^2+1}\iff a^2=\dfrac{2+x}{1-x}$

$$\iff\cos2t=\dfrac{1-a^2}{1+a^2}=\dfrac{1-x-(2+x)}{1-x+2+x}=-\dfrac{3+2x}3$$

Use $$\sin^22t+\cos^22t=1$$

or without Trigonometry,

$$\left(\dfrac{2a}{1+a^2}\right)^2+\left(\dfrac{1-a^2}{1+a^2}\right)^2=1$$

Answer 2

Hint:

You can do it in two ways:

• eliminate $a$. For instance solve the first equation for $a^2$ and substitue in the second equation.

• evaluate $(x-\frac12)^2+y^2$.

Answer 3

From the first equation we have $$a^2=\frac{x+2}{1-x}$$ and $$(x+2)^2+y^2=\left(\frac{3a^2}{a^2+1}\right)^2+\frac{9a^2}{(a^2+1)^2}=\frac{9a^2}{a^2+1}=3(x+2),$$ which gives which you wish.