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 $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$


Hint At least for $(a, b) \neq (-1, 0)$, which is not realized by any value $t$, draw the line through $(-1, 0)$ and $(a, b)$ in the $xy$-plane. Writing the point $(a, b)$ of intersection of the line and the unit circle $x^2 + y^2 = 1$ in terms of the slope $t$ of the line gives exactly $$(a, b) = \left(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2}\right),$$ so the claim is true for all $(a, b) \neq (-1, 0)$. On the other hand, substituting shows that there is no value $t$ for which $(a, b) = (-1, 0)$.
