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: Complete the square for \(x\) terms

Group \(x\) terms: \(x^{2}-12x=(x - 6)^{2}-36\) (using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2b=12\Rightarrow b = 6\))

Step2: Complete the square for \(y\) terms

Group \(y\) terms: \(y^{2}-2y=(y - 1)^{2}-1\) (using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2b = 2\Rightarrow b=1\))

Step3: Rewrite the circle equation

The original equation \(x^{2}+y^{2}-12x - 2y+12 = 0\) becomes \((x - 6)^{2}-36+(y - 1)^{2}-1+12=0\)
Simplify to \((x - 6)^{2}+(y - 1)^{2}=25\)
The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center.

Answer:

\((6,1)\)