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

centre = (6, 1 )

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + y² - 12x - 2y + 12 = 0

Collect the x and y terms together and subtract 12 from both sides

x² - 12x + y² - 2y = - 12

To obtain the equation in standard form use completing the square

add ( half the coefficient of the x/ y term )² to both sides

x² + 2(- 6)x + 36 + y² + 2(- 1)y + 1 = - 12 + 36 + 1

(x - 6)² + (y - 1)² = 25 ← in standard form

with centre = (6, 1 ) and radius = = 5

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation: