If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:

If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$


The easiest way is by brute checking.
\begin{align*}
a^2&=\frac{(t^2-1)^2}{(t^2+1)^2}\\
b^2&=\frac{4t^2}{(t^2+1)^2}\\
a^2+b^2&=\frac{t^4-2t^2+1+4t^2}{(t^2+1)^2}\\
&=\frac{t^4+2t^2+1}{(t^2+1)^2}\\
&=\frac{(t^2+1)^2}{(t^2+1)^2}\\
&=1
\end{align*}
So we know that if we can find a $t$ everything works out fine. Now observe that if $t=\frac{b}{a+1}$ the two expressions for $a$ and $b$ in terms of $t$ hold, so for any point in the circle other than $(-1,0)$ this works.
