To determine the center of the circle given by the equation [tex]\( x^2 + y^2 - 12x - 2y + 12 = 0 \)[/tex], we can follow these steps to transform the equation into its standard form, which is [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. Here, [tex]\((h, k)\)[/tex] is the center of the circle, and [tex]\(r\)[/tex] is its radius.
1. Rewrite the equation grouping [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[
x^2 + y^2 - 12x - 2y + 12 = 0
\][/tex]
2. Move the constant term to the other side:
[tex]\[
x^2 - 12x + y^2 - 2y = -12
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[
x^2 - 12x = (x - 6)^2 - 36
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[
y^2 - 2y = (y - 1)^2 - 1
\][/tex]
5. Substitute back into the equation:
[tex]\[
(x - 6)^2 - 36 + (y - 1)^2 - 1 = -12
\][/tex]
6. Combine constants on one side:
[tex]\[
(x - 6)^2 + (y - 1)^2 - 37 = -12
\][/tex]
7. Isolate the squared terms:
[tex]\[
(x - 6)^2 + (y - 1)^2 = 25
\][/tex]
Now that we have the equation in the standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex], we can identify the center [tex]\((h, k)\)[/tex]:
From [tex]\( (x - 6)^2 + (y - 1)^2 = 25 \)[/tex],
The center [tex]\((h, k)\)[/tex] is [tex]\((6, 1)\)[/tex].
Therefore, the center of the circle is [tex]\(\boxed{(6, 1)}\)[/tex].
