Answer:
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
