It is possible to compute the Picard group of (some? all?) genus $0$ curves in the following manner:
For concreteness let, $A = k[x,y]/(x^2+y^2-1)$ and $X = \operatorname{Spec} A$. Let $Y$ denote $\operatorname{Proj} k[x,y,z]/(x^2+y^2-z^2)$. Let $k$ be any field for now.
Then, either by general theory or by explicit computation, we can find an isomorphism $\pi: Y \to \mathbb P^1$. Now $X$ is an open subscheme of $Y$ that either misses two degree one points or one degree two points depending on whether $k$ contains $\sqrt{-1}$.
By the above isomorphism, we know that $\operatorname{Pic} Y = \operatorname{Pic} \Bbb P^1 = \Bbb Z$. The Picard group is easily seen to be isomorphic to the Class groups in this case and the excision sequence gives us the following: $$G \to \operatorname{Cl} Y \to \operatorname{Cl} X \to 0$$ where $G = \Bbb Z$ or $\Bbb Z\oplus \Bbb Z $ depending on whether $X$ misses one point or two. This lets us compute the class group of $X$ and it turns out to be either trivial or $\Bbb Z/2$.
The analogy:
I believe there is some sort of strong analogy between $A$ and $\Bbb Z[\sqrt{-5}]$ when $k$ does not contain $\sqrt{-1}$. For instance, I think both rings are not UFD's "only" because of the factorizations: $$x^2 = (1-y)(1+y), 2\times 3 = (1+\sqrt{-5})(1+\sqrt{-5})$$ where $x$ is roughly equivalent to $2/3$ and $y$ is $\sqrt{-5}$. Both factorizations seem to occur because we can write a square as a difference of two squares($5 = 3^2-2^2$.)
Questions:
Can one make this analogy precise? Does it extend to other genus $0$ curves/(quadratic?) number fields? What about higher genus?
If one can make this analogy precise, does the above method also let us compute the Class groups of number fields?
A reference(with maybe a short explanation if possible) would be perfectly acceptable as an answer.
Usually the analogy is drawn between number fields and curves over finite fields, concerning such vast and deep topics as CFT or L-functions. However you appear to ask for more than analogy in the specific case of genus zero. As stressed by Captain Lama, "computing the class group of a number field is something so ridiculously difficult" that your question looks like wishful thinking. But if you accept to abandon the « global » point of view for the « local » one, i.e. to look at phenomena at one single prime at a time, there may be more than an analogy. More precisely, fix a number field K and a prime number p (suppose p odd for simplicity), and denote by S the set of primes above p in K. Let $G_S(K)$ be the Galois group over K of its maximal pro-p-extension which is unramified outside S. One would like to give a description of $G_S(K)$ by generators and relations in the category of pro-p-groups. A minimal set of generators of $G_S(K)$ can be obtained using CFT and, roughly speaking, the $Z_p$-torsion submodule $T_S(K)$ of the abelianization of $G_S(K)$ « contains » the relations. The field K is called p-rational if $T_S(K)$ is zero, i.e. if $G_S(K)$ is pro-p-free. This is the exact analogue of the case of a curve X (projective, smooth, irreducible) over an algebraically closed field, where the role of $G_S(K)$ is played by the profinite fundamental group of X \ {${P_1, … , P_s , infty}$}, s being the cardinal of S . For details and developments, see Movahhedi & Nguyen Quang Do, « Sur l’arithmétique des corps de nombres p-rationnels », in Séminaire de Théorie des Nombres Paris 1987-88, Progress in Math., vol. 81, Birhäuser, especially pp. 157-158. The condition of p-rationality imposes of course conditions on the p-class group. If K is a quadratic imaginary field and p is at least 5, K is p-rational if p does not divide theclass number of K. This is the case of your example, but unfortunately, the information goes in the wrong sense !