Answer:
(0, 1)
Step-by-step explanation:
A, B, and C are points on a circle. You are being asked to find the center of the circle. You can do that any of several ways.
I find it useful to plot the points on a graph. Doing so reveals that AB is perpendicular to BC, so ABC is a right triangle, and the center you want is the midpoint of the hypotenuse (AC).
(A+C)/2 = ((2, -3) +(-2, 5))/2 = (2-2, -3+5)/2 = (0, 1) . . . . . coordinates of center
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From a construction point of view, the center of the circle is the point of intersection of the perpendicular bisectors of the chords. In the attached, we have used a geometry program to draw the perpendicular bisectors of AB and BC. They meet at (0, 1).
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From an algebra point of view, the points on the circle with center (h, k) and radius r satisfy the equation ...
(x -h)^2 +(y -k)^2 = r^2
Filling in the point values, you get three equations in the three unknowns (h, k, r):
- (2-h)^2 +(-3-k)^2 = r^2
- (4-h)^2 +(3-k)^2 = r^2
- (-2-h)^2 +(5-k)^2 = r^2
Finding the difference between any two pairs of these will result in two linear equations in h and k, so are easily solved.
First - second: (4 -4h +h^2) +(9 +6k +k^2) -(16 -8h +h^2) -(9 -6k +k^2) = r^2 -r^2
-12 +4h +12k = 0 . . . simplify
h +3k = 3 . . . . . . . . . put in standard form
Second -third: (16 -8h +h^2) +(9 -6k +k^2) -(4 +4h +h^2) -(25 -10k +k^2) = r^2 -r^2
-4 -12h +4k = 0 . . . . simplify
3h -k = -1 . . . . . . . . . put in standard form
Now, we can add 3 times the second of these equations to the first to find h:
(h +3k) +3(3h -k) = (3) +3(-1)
10h = 0
h = 0
Then either of the equations can be used to find k = 1.
The center is (h, k) = (0, 1).
