What is the center of a circle whose equation is x² + y² - 12x - 2y + 12 = 0?

Solution:

Given equation is x² + y² -12x - 2y + 12 = 0

Comparing with the general equation of a circle, x² + y² + 2gx + 2fy + c = 0,

2g = -12, 2f = -2, c = 12

g = -6, f = -1

Centre = (-g, -f)= (6, 1)

Given equation is x² + y² - 12x - 2y + 12 = 0

By completing the square,

x² - 12x + 36 + y² - 2y + 1 = -12 + 36 + 1

(x - 6)² + (y - 1)² = 25

(x - 6)² + (y -1)² = 5²

Comparing with equation of a circle (x - h)²+ (y - k)² = r² centred at (h, k) and radius 'r',

Centre = (h, k) = (6,1)

What is the center of a circle whose equation is x² + y² - 12x - 2y + 12 = 0?

Summary:

The center of the circle of the given equation is x² + y² - 12x - 2y + 12 = 0 is (6,1).