I've been trying to solve the next problem but I can't complete the last step.
The problem is:
Considering the image of the circle $$V=\{ (x_1,x_2): x_1^2+x_2^2=1 \}$$ given by the map $$\rho: V \rightarrow \mathbb{A}^2(\mathbb{R})$$ $$(x_1,x_2) \longmapsto (\frac{x_1}{1+x_2^2}, \frac{x_1x_2}{1+x_2^2})$$ Compute the equations of the image.
My attemp: The map $\rho$ is well defined over the open set $$U=\{(x_1,x_2)\in V: g=(1+x_2^2)(1+x_2^2) \neq 0\}=V$$ so we can take the affine variety $ Y:=\mathbb{A}^2(\mathbb{R})_g \subset \mathbb{A}^3(\mathbb{R})$ given by $$Y:=\{(x_1,x_2,z):zg(x_1,x_2)-1=0\}=\{(x_1,x_2,z):z(1+x_2^2)^2-1=0\}$$ and a morphism $\phi: Y \rightarrow W $ such that $\phi(Y)= \rho(V)$. $$ $$
Calling $ \Gamma_\phi$ to the graph of $\phi$ in $ \mathbb{A}^3(\mathbb{R}) \times \mathbb{A}^2(\mathbb{R})$, we have that: $$I(\Gamma_\phi)=\langle z(1+x_2^2)^2-1, y_1-x_1 (1+x_2^2), y_2-x_1x_2(1+x_2^2) \rangle$$
Now we have to compute a Gröbner basis of that ideal and, after that, use elimination theory to quit all the expresions in that basis that depend on $x_1,x_2,z$. The remaining equations will be those asked in our problem. I've tried this last part many times but I always get all the equations depending on $x_1,x_2,z$. I'd really appreciate if someone could help me in this last part. Thanks in advance.
Solving equations $y_1=\frac{x_1}{1+x_2^2}$ and $y_2= \frac{x_1x_2}{1+x_2^2}$ for $x_1$ and $x_2$ and substituting into $x_1^2+x_2^2=1$, I get $y_1^3+y_2^2y_1+y_2^2=y_1^2$. Does that help?