We don't need to do it like the other tutor suggests. We don't even need to
know the equation of a circle.  You may not have even studied that yet,
anyway.  We just plot those points and draw three circles with diameter 10
(radius 5) with centers (0,0), (5,5) and (10,5):



So we just look and see that a circle with a diameter of 10 units and 
passes through the coordinates A,C,D, and E, and all three circles 
have centers with integer coordinates.

So if one of those points can't lie on a circle's boundary
with integer coefficients and radius 5, it would have to be B(0,0).

We can quit here and figure that the answer can only be B(0,0).

--------------------------

However we haven't really shown that B(0,0) can't lie on a circle's boundary
with integer coefficients and radius 5.  We've only gotten it by elimination.

Let's draw a circle (in red) with radius 5 that goes through B(0,0)
and also through (5,0).  Let's also draw in two radii and label 
the center O. 



That forms an equilateral triangle, because all sides are 5 units long.
                                                               _
Since we know that the altitude of an equilateral triangle is √3/2 times
the length of a side of the equilateral_triangle, we know that the y-
coordinate of the center O is 5 times √3/2, and is irrational.

And it's the same with the only other circle we could draw with radius
5 that goes through B(0,0) and (5,0). 

 

Edwin