The diagonal BD has equation y=1/2x+3.
Then I know that the angLe between the diagonal of the square and one of the sides is 45.
tan(45+a)=(1+tan a)/(1-tan a) is the gradient of the line AD where a= angle between the diagonal and the x-axis.
So the gradient of AD is 3, the line equation is y= 3x+18 and D(0,-6)
For AB the gradient is just -1/3 so I get y=-1/3x -14/3 and B(2,4)
BC has gradient 3, y= 3X-2, CD y=-1/3x -2 and C(0,-2).
To get the centre you need only to find that line AC is y=-2x-2. Then the centre O is (-2,2).
For the centre of the circle, i remember that for three points which aren't on the same line you can always draw a circle, by first drawing the axis of the two segment formed by joining the points. So if you find the equations of the segment OA which is y=0 and has its middle point at x= 1.
And then the segment OB y= 2x and has the middle point at (2,4).
Find the lines which passes trhough these points and are perpendicular to the two segments,
x=1 and y= -1/2x +5 and then find the intersection which is (1,9/2).
I don't know if it is right, I hope it helps.