To determine the center of the circle represented by the equation x² + y² – 12x – 2y + 12 = 0, we first need to rewrite this equation in the standard form of a circle equation:
The standard form of a circle’s equation is given as:
(x – h)² + (y – k)² = r²
where (h, k) is the center of the circle and r is the radius.
We start by rearranging the given equation:
x² - 12x + y² - 2y + 12 = 0
Next, we group the x and y terms:
(x² - 12x) + (y² - 2y) + 12 = 0
To complete the square, we need to add and subtract the appropriate values for both the x and y terms:
- For the x terms:
– The coefficient of x is -12.
– Half of -12 is -6, and squaring it gives 36. - For the y terms:
– The coefficient of y is -2.
– Half of -2 is -1, and squaring it gives 1.
Now we add 36 and 1 to both sides of the equation:
(x² - 12x + 36) + (y² - 2y + 1) + 12 - 36 - 1 = 0
This simplifies to:
(x - 6)² + (y - 1)² - 25 = 0
Now, we can rearrange it to the standard form:
(x - 6)² + (y - 1)² = 25
From this final equation, we can identify the center of the circle:
- h = 6
- k = 1
Thus, the center of the circle is located at the point:
(6, 1)
In summary, the center of the circle defined by the equation x² + y² – 12x – 2y + 12 = 0 is the point (6, 1).