Finding the point through which a circle passes if the general equation is given No idea how it is even possible to do this. 

So we have the general equation (but the y-coordinate of the center is still kept variable as you can see below):-
$$x^2+y^2-2x-3ky-2=0$$

We have to find through which two fixed points does the circle pass.
Hint If the points $(x, y)$ of interest satisfy the equation for distinct parameter values $k$ and $k'$, then subtracting the two equations gives that $(x, y)$ must satisfy $$3ky' - 3ky = 0.$$
Hint: at which points does your circle meet the axes?