The Missing Intercept

I posted this puzzle some time ago,

but no one provided a satisfactory answer.

Given: .$\displaystyle \begin{Bmatrix}x \:=\:\dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y \:=\:\dfrac{2t}{1+t^2} \end{Bmatrix}$

We have the parametric equations of a unit circle.

Verification

Square: .$\displaystyle \begin{Bmatrix}x^2 \:=\:\dfrac{(1-t^2)^2}{(1+t^2)^2} & [1]\\ \\[-3mm] y^2 \:=\:\dfrac{(2t)^2}{(1+t^2)^2} & [2]\end{Bmatrix}$

Add [1] and [2]: .$\displaystyle x^2+y^2\:=\:\frac{1 - 2t^2 + t^4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2} $

$\displaystyle = \;\frac{1 + 2t^2+t^4}{(1+t^2)^2} \:=\:\frac{(1+t^2)^2}{(1+t^2)^2} \;=\;1 $

Hence, a unit circle: .$\displaystyle x^2+y^2\:=\:1$

To find the $\displaystyle y$-intercepts, let $\displaystyle x = 0.$

$\displaystyle x = 0\!:\;\;\frac{1-t^2}{1+t^2}\:=\:0 \quad\Rightarrow\quad t \:=\:\pm1$

Hence, the $\displaystyle y$-intercepts are: .$\displaystyle (0,1),\;(0,\text{-}1)$

To find the $\displaystyle x$-intercepts, let $\displaystyle y = 0.$

$\displaystyle y = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0 \quad\Rightarrow\quad t \:=\:0$

Hence, the $\displaystyle x$-intercept is: .$\displaystyle (1,0)$ ?

Where is the other $\displaystyle x$-intercept?