What is $I(X)$ for $X\subseteq\mathbb{A}^2$ given by $x^2+y^2=x=1$?

Question

Suppose $k$ is algebraically closed, $\mathbb{A}^2$ affine. I'm curious about the subset $X\subseteq\mathbb{A}^2$ given by the unit circle $x^2+y^2=1$ and the line $x=1$. What would the ideal $I(X)$ be in this case?

I have $f\in I(X)$ if $f(x,y)=0$ for $(x,y)\in X$. Then $g(y)=f(1,y)=0$ for all $y$, so $f$ would be a product of $y$ and a linear factor? Testing points on $x^2+y^2=1$ started giving me strange things so I think this might be incorrect.

Answer 1

Well, $x = 1$ and $x^2 + y^2 = 1$ give $y^2 = 0$. Thus, the only point in $X$ is $x = 1, y = 0$. This point corresponds to the maximal ideal $(x - 1, y)$.

Answer 2

The ideal will be the following:

$$ \langle \; x^2 + y^2 - 1,\; x-1 \; \rangle = \langle \; y^2,\; x-1 \; \rangle $$

which corresponds to the point $(1,0)$.

Here with $\langle a, b \rangle$ I mean the ideal generated by $a$ and $b$. Notice that for the second step you basically use $x=1$.