Show that $mathfrak{m}_p$ is an ideal in $mathcal{O}_V.$

I’m working through Algebraic Geometry: A Problem Solving Approach by Garrity et al, and I have found myself stuck on Exercise 4.13.1, which is the section Points and Local Rings.

Let $V = V(x^2 + y^2 – 1) subset mathbb{A}^2(k).$ Let $p = (1, 0)
in V.$ Define $$mathfrak{m}_p = {f in mathcal{O}_V : f(p) =
0}.$$

Show that $mathfrak{m}_p$ is a maximal ideal in
$mathcal{O}_V.$

We’re told that $mathcal{O}_V$ is a subring of the function field $mathcal{K}_V = {frac{f}{g} : f, g in mathcal{O}(V), g neq 0},$ where $mathcal{O}(V) = k[x, y]/I(V).$

It was easy enough to show that $mathfrak{m}_p$ is an ideal, but I’m getting caught up on how to show that it is maximal.

I know that maximal ideals in $k[x, y]$ are all of the form $I = langle x – a_1, y – a_2 rangle,$ so my idea was to show that $mathfrak{m}_p = langle x – 1, y rangle,$ and then I know that maximal ideals in $k[x, y]/I(V)$ correspond to maximal ideal in $k[x, y]$ containing $I(V).$ This feels close to something correct, but I think I just don’t understand the different spaces well enough to put it all together. Also, $mathfrak{m}_p$ is an ideal in $mathcal{O}_V,$ while $langle x – 1, yrangle$ is an ideal in $k[x, y],$ so setting them equal to each other doesn’t make any sense.