The origin of a mixed chart lies in the Zariski closure of complex points of the chart curve with $u\ne0$

Let $m\ge1$ be an integer, let $a,b\in\mathbb C$ be arbitrary, and let $Q_{a,b}(u,y)=a\,u+y+b\,u^{m+1}y^m+u^my^{m+1}\in\mathbb C[u,y]$. With $P_\alpha(X,Y)=X^m(\alpha+XY)+Y^m(\bar\alpha+XY)$, the choice $(a,b)=(\alpha,\bar\alpha)$ gives the polynomial equal to $u^{m+1}P_\alpha(1/u,y)$ for $u\ne0$, the equation of the curve $P_\alpha=0$ in the chart $u=1/X$ near the boundary point $(\infty,0)$ of $\mathbb P^1\times\mathbb P^1$; the choice $(a,b)=(\bar\alpha,\alpha)$ gives the analogous equation near $(0,\infty)$, with the two coordinates exchanged. For $v\in\mathbb C^2$ let $\mathfrak m_v$ be the maximal ideal of the polynomials vanishing at $v$, a point of $\operatorname{Spec}\mathbb C[u,y]$.

Then, with the closure taken in the Zariski topology of $\operatorname{Spec}\mathbb C[u,y]$,

The approximating points are complex points of the chart curve at which $X=1/v_0$ is finite; for $(a,b)=(\alpha,\bar\alpha)$ they correspond to the points $(1/v_0,v_1)$ of the affine curve $P_\alpha=0$. So the corner represented by the origin lies in the Zariski closure of points of the affine curve, as asserted for the closure of the curve (6) in $\mathbb P^1\times\mathbb P^1$.

Formalization Note: points of $\operatorname{Spec}\mathbb C[u,y]$ are elements of PrimeSpectrum (MvPolynomial (Fin 2) ℂ) with Mathlib's Zariski topology, $u$ and $y$ are the variables X 0 and X 1, and $\mathfrak m_v$ is the kernel of evaluation at $v$.