Question

what is the center of a circle whose equation is $x^{2}+y^{2}-12x-2y+12=0$?
$(-12, -2)$
$(-6, -1)$
$(6, 1)$
$(12, 2)$

Explanation:

Step1: Rearrange the equation

$x^2 - 12x + y^2 - 2y = -12$

Step2: Complete the square for x

Take half of -12, square it: $(-6)^2=36$. Add to both sides.
$x^2 -12x +36 + y^2 -2y = -12 +36$

Step3: Complete the square for y

Take half of -2, square it: $(-1)^2=1$. Add to both sides.
$x^2 -12x +36 + y^2 -2y +1 = -12 +36 +1$

Step4: Rewrite in standard form

Standard circle form: $(x-h)^2+(y-k)^2=r^2$, center $(h,k)$.
$(x-6)^2 + (y-1)^2 = 25$

Answer:

(6, 1)