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

The center of this circle is at (6, 1).

Step-by-step explanation:

Rewrite x^2+y^2-12x-2y+12=0 by grouping like terms.  Then:

x^2+y^2-12x-2y+12=0 becomes x^2 - 12x +y^2 - 2y + 12=0.

Next, complete the squares:

x^2 - 12x + 36 - 36 + y^2 - 2y + 1 - 1 + 12 = 0.

Rewriting the two perfect squares as squares of binomials, we get:

(x - 6)^2 - 36 + (y - 1)^2 - 1 + 12 = 0

Moving the constants to the right side:

(x - 6)^2     + (y - 1)^2   = 36 + 1 - 12 = 25

Then the desired equation is:

(x - 6)^2 + (y - 1)^2 = 5^2.  The center of this circle is at (6, 1).

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step-by-step explanation:

da freeck kinda math is this just drop out

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