Answer
The equation of the line of slope $t$ through $(-1,0)$ is given by
(1)${\ }$ $y=t (x+1)$.
Work Step by Step
The equation of the line of slope $t$ through $(-1,0)$ is given by
$y-0=t (x-(-1))$,
so,
(1)${\ }$ $y=t (x+1)$.
We have the equation of the unit circle $x^2+y^2=1$. Substituting $y$ from Eq. (1) into $x^2+y^2=1$ gives
$x^2+t^2(x+1)^2=1$,
$x^2+t^2 (x^2+2x+1)=1$,
$x^2+t^2 x^2+2t^2 x +t^2=1$,
$(1+t^2)x^2+2t^2 x+(t^2-1)=0$.
Now, this a quadratic equation of the form: $a x^2+b x+c=0$. The solutions are
$x=\frac{-b\pm \sqrt{b^2-4 a c}}{2 a}$. So, solving for $x$ we obtain
$x=-1$ or $x=\frac{1-t^2}{t^2+1}$.
Substituting $x=-1$ into (1) we get $y=0$. Substituting $x=\frac{1-t^2}{t^2+1}$ into (1) we get $y=\frac{2 t}{t^2+1}$.
The solution $x=-1$ is just the original point $(-1,0)$. So, the line of slope $t$ through $(-1,0)$ intersects the unit circle in the point with coordinates $(x,y)=\left(\frac{1-t^2}{t^2+1},\frac{2 t}{t^2+1}\right)$. Since the slope $t$ is arbitrary, these coordinates determine all points on the unit circle except the point $(-1,0)$, we conclude that the unit circle has parametrization $\left(\frac{1-t^2}{t^2+1},\frac{2 t}{t^2+1}\right)$ excluding the point $(-1,0)$.
Furthermore, from equation (1) we obtain $t=y/(x+1)$.
