In this blog of Alex Youcis, I see a sentence in the proof of theorem 4 which says that "since $C$ has genus 0 that it defines a class $[C]\in\mathrm{Br}(\mathbb{Q})$ which is trivial if and only if $C\cong\mathbb{P}^1_\mathbb{Q}$". And the definition of the Brauer group of a field is the equivance classes of central simple algebras.
So the question is that what the map and the inverse map are between $\mathrm{Br}(\mathbb{Q})$ and the set of all smooth curves of genus $0$ over $\mathbb{Q}$(I don't know if this is a set).
My thoughts: Smooth curves of genus $0$ corresponds to a quadratic form over $\mathbb{Q}$(By Riemann-Roch and embed this curve into $\mathbb{P}^2$ as a quadratic curve), and a quadratic form corresponds to an element in the Brauer group. Is this right?
Thanks!
Just to get this off the unanswered list I will turn my comment in to an answer.
In general for a field $k$ there is a bijection between (isomorphism classes of) central simple algebras over $k$ of dimension $n^2$ and twists of $\mathbb{P}^{n-1}_k$. One can find the details of this in Bjorn Poonen's book Rational points on varieties or Gille and Szamuely's book Central simple algebras and Galois cohomology. The operative phrases here are 'Brauer group', 'central simple algebras', and 'Brauer-Severi varieties'.
As you guessed, under this correspondence the conic $V(ax^2+by^2-z^2)\subseteq\mathbb{P}^2_k$, which is a twist of $\mathbb{P}^1_k$, corresponds to the quaternion algebra $\left(\frac{a,b}{k}\right)$.