What is the center of a circle circumscribed about a triangle with vertices (−2, 2) ( 2, −2)and (6, 2)?

General equation for circle of radius r with center (h,k):
 (x-h)² + (y-k)² = r²

(-2,2) and (6,2) are horizontally aligned.
The vertical line x=2 passes halfway between them. The center of the circle lies on this line.
(x-2)² + (y-k)² = r²

Plug two points into (x-2)² + (y-k)² = r² and solve the system for k.
(-2,2): k² - 4k = r² - 20
(2,-2): k² + 4k = r² - 4
k = 2

(x-2)² + (y-2)² = 16
center (2,2)

Construct perpendicular bisectors of at least 2 of the sides of the triangle and see where they intersect

Points: (-2 , 2) , (2 , -2)
m = (-2 - 2) / (2 - (-2)) = -4 / 4 = -1
Perpendicular slope: 1
midpoint: (-2 + 2) / 2 = 0 , (2 + (-2)) / 2 = 0 =>> (0 , 0)

y - 0 = 1 * (x - 0)
y = x

Points: (2 , -2) and (6 , 2)
m = (2 - (-2)) / (6 - 2) = 4/4 = 1
Perpendicular slope = -1
midpoint: (2 + 6) / 2 = 4 , (-2 + 2) / 2 = 0 (4 , 0)

y - 0 = -1 * (x - 4)
y = 4 - x


y = x
y = 4 - x

x = 4 - x
2x = 4
x = 2

y = x
y = 2

(2 , 2) is the center of the circle