This is really a comment but will be too long and, anyway, I'm not entitled. There is a standard approach to this problem but I haven't carried through the computations. Let me indicate the method for euclidean case first. Suppose we have points $A_1,\dots,A_6$ in the plane, with the first five fixed, say $A_1=(a_{11},a_{12})$ etc. but $A_6=(x,y)$ variable. Now compute $X,Y,Z$,the intersections of $A_2 A_6$ and $A_5A_3$, etc. Then apply the condition for these three points to be collinear. This turns out to be a quadratic in $x$ and $y$ which proves Pascal's theorem AND its converse. These computations are rather tedious but can be carried in a few minutes with, say, Mathematica. They can even be done by hand by assuming wlog that three of the original points are $(0,0),(1,0),(0,1)$.

This is really a comment but will be too long and, anyway, I'm not entitled. There is a standard approach to this problem but I haven't carried through the computations. Let me indicate the method for euclidean case first. Suppose we have points $A_1,\dots,A_6$ in the plane, with the first five fixed, say $A_1=(a_{11},a_{12})$ etc. but $A_6=(x,y)$ variable. Now compute $X,Y,Z$,the intersections of $A_2 A_6$ and $A_5A_3$, etc. Then apply the condition for these three points to be collinear. This turns out to be a quadratic in $x$ and $y$ which proves Pascal's theorem AND its converse. These computations are rather tedious but can be carried in a few minutes with, say, Mathematica. They can even be done by hand by assuming wlog that three of the original points are $(0,0),(1,0),(0,1)$.

The same programme can be carried in your case---you need the formula for the intersections of the lines through two pairs of points and the condition for three points to be collinear in spherical geometry. Presumably, these are available somewhere in the literature. As I said, I haven't done this explicitly so can't tell which class of curves solves your problem.