I'm given the coordinates of two vertices of a square: $(5,1)$ and $(8,5)$ and I'm asked to find the other two.
I suppose there are two squares which meet these conditions. I can obtain a point $(x,y)$ by solving two equation derived from the distance formula for the aforementioned points. One equation implies that the side length of the square is $5$ and the other implies that the diagonal is $5 \sqrt{2}$
$$x^2-16x+y^2-10y+64=0$$
$$x^2-10x+y^2-2y-24=0$$
Solving these equations would give me two points each in a different square, and I can do the same for the other two points. But this approach is tiring and time-consuming. Is there a simpler approach?
First translate the problem to be a little simpler.
Now there is one point at $(0,0)$ and the other is at $(3,4)$. So if that is one side of the square, you can put the other two to the upper left. You need one of them to be a distance 5 from the origin and give a perpendicular edge so it should be at $(-4,3)$. Now fill in the top corner and translate back.
If this is the diagonal of the square you have that the sides are distance $5/\sqrt{2}$ so the other two are given by the intersections of two circles. The first one being $x^2 + y^2 = 50$ and the second being $(x-3)^2 + (y-4)^2 = 50$. But this is the procedure you already put, but it should be simpler now that you've gotten rid of some of the extraneous numbers by translation.