The center of the circumscribed circle O around triangle ABC with vertices A(0,0), B(12,7), and C(12,0) can be found by locating the midpoint of side AB and understanding that it lies on the horizontal line y=0. The coordinates are (6, 3.5).
To find the center of circle O that is circumscribed around triangle ABC with vertices A(0,0), B(12,7), and C(12,0), we need to find the intersection of the perpendicular bisectors of at least two sides of the triangle.
The side AB can be bisected by finding its midpoint and then finding the slope of a line perpendicular to AB. The midpoint of AB is ( (0+12)/2, (0+7)/2 )=(6, 3.5). The slope of AB is (7-0)/(12-0)=7/12, so the slope of the line perpendicular to AB is -12/7. Similarly, for side AC, the midpoint is ( (0+12)/2, (0+0)/2 )=(6, 0), and since AC is a vertical line, the perpendicular bisector of AC must be a horizontal line running through the midpoint of AC.
Therefore, the center of the circle O must lie on the horizontal line y=0 at x=6, which is also the x-coordinate of the midpoint of AB. Since we are looking for the intersection of the perpendicular bisectors, and we know one bisector is the horizontal line at y=0, the center of the circle will be at the midpoint of AB. So the coordinates of the center of the circle O are (6, 3.5).