What is the center of a circle whose question is x2+y2-12x-2y+12=0?

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

Step-by-step explanation:

Hint:  for clarity please use the " ^ " symbol to denote exponentiation.

We have:  x^2+y^2-12x-2y+12=0

and should separate the x and y terms, as follows:

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

For each x and y, we must now "complete the square."  

Focusing first on x^2 - 12x, take half of the coefficient of x, which here is -12, and then square the result:  half of -12 is -6, and the square of -6 is +36.

Add 36 to x^2 - 12x and then subtract 36, obtaining:

x^2 - 12x + 36 - 36

Now rewrite x^2 - 12x + 36 as a perfect square:  (x - 6)^2.

Then our x^2 - 12x becomes (x - 6)^2 - 36.

Similarly, our y^2 - 2y becomes (y - 1)^2 - 1.

Overall, we then have:

(x - 6)^2 - 36 +  (y - 1)^2 - 1 = -12

Add 36 + 1 to the left side and also to the right side.  This results in:

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

(x - 6)^2 + (y -1)^2 = 25, and 25 = 5^2.

Thus, the standard equation of this particular circle is

(x - h)^2 + (y - k)^2 = r^2

Comparing this to what we obtained earlier, we see that h = 6, k = 1 and r = 5.

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

(x - 6)^2 + (y -1)^2  = 5^2

Comparing this to

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