Given that the equation of circle C is x ^ 2 + (Y-1) ^ 2 = 4, circle C and circle C with the center (- 2,1) intersect at two points a and B, and | ab | = 2 root sign 2, the equation of circle C is solved
Circle C: x ^ 2 + (Y-1) ^ 2 = 4
Let C ': (x + 2) ^ 2 + (Y-1) ^ 2 = R
Center distance d = √ (0 + 2) ^ 2 + (1-1) ^ 2 = 2
The center distance is the sum of the distance between the center of two circles and the midpoint of the common chord, and the common chord is perpendicular to the center line
∴√(4-2)+√(r^2-2)=2
The solution is R ^ 2 = 8-4 √ 2
Ψ circle C ': (x + 2) ^ 2 + (Y-1) ^ 2 = 8-4 √ 2