Answer:
17
Explanation:
The geometry program that drew the circle through the three given points wrote this equation for the circle:
(x -6)^2 +(y -11)^2 = 130
Compared to the standard form equation, ...
(x -h)^2 +(y -k)^2 = r^2 . . . . . . circle with center (h, k) and radius r
we can see that the center is (h, k) = (6, 11).
The sum h+k is 6+11 = 17.
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Additional comment
The center is at the point of intersection of the perpendicular bisectors of chords of the circle.
Here it is convenient to choose one of them as the horizontal segment with endpoints (3, 0) and (9, 0). The vertical line that bisects this segment is x=(3+9)/2 = 6. So, x=6 is the x-coordinate of the circle center.
We can use endpoints (-1, 2) and (3, 0) as another chord. Its midpoint is (-1+3, 2+0)/2 = (1, 1). The slope of this chord is (y2-y1)/(x2-x1) = (0-2)/(3-(-1)) = -1/2. So, the slope of the perpendicular bisector is -1/(-1/2) = 2, and the equation of that line is
y -1 = 2(x -1)
For x=6, the point of intersection of the chord bisectors is ...
y -1 = 2(6 -1) = 10
y = 11
The coordinates of the circle center are (6, 11), so h+k = 6+11 = 17.