The center of the circle of that passes through the points (-2,3), (0,-3), and (4,-3) is ________.?

Answer 1

In general, 3 points will determine a circle. Its center is the point of intersection of the perpendicular bisectors of the chords formed by the 3 points.

Notice that (0,-3) and (4,-3) form a horizontal chord, so we know the center is somewhere on the vertical line x = 2.

The chord (-2,3) -- (0,-3) has a slope of -3 and a midpoint at (-1,0), so its perpendicular bisector has a slope of 1/3, its equation is y-0 = (1/3)(x+1), and it intersects x=2 at y = (1/3)(2+1) = 1.

So the center is (2,1).

Just to be sure, distance from (2,1) to (-2,3) is √(16 + 4) = √20; from (2,1) to (0,-3) is √(4 + 16) = √20; from (2,1) to (4,-3) is √(4 + 16) = √20. So it's definitely the center.

Answer 2

The center of the circle is the point that is equidistant from all three points. It must lie on the perpendicular bisector of the segment with endpoints (-2,3) and (0,-3) and also on the perpendicular bisector of the segment with endpoints (0,-3) and (4,-3). Since the second segment is horizontal, we need the equation of the vertical line through its midpoint (2,-3)which is x=2. The midpoint of the first segment is (-1,0). The slope of the first segment is -3 so the slope of the perpendicular to the first segment is 1/3. The equation of its perpendicular bisector i.e. the line through (-1,0) with slope 1/3 is y=(1/3)(x+1). This line crosses the vertical line x=2 at point (2,1) which is the center of the circle through the three points.