To find the center of a circle from its equation, we need to rewrite the equation in the standard form of a circle, which is:
(x – h)² + (y – k)² = r²
where (h, k) is the center of the circle and r is the radius.
Given the equation:
x² + y² – 12x – 2y + 12 = 0
First, let’s rearrange this equation:
x² – 12x + y² – 2y + 12 = 0
Next, we move the constant term to the other side:
x² – 12x + y² – 2y = -12
Now we apply the method of completing the square for the x and y terms.
1. For the x terms: x² – 12x
– Take half of -12, which is -6, and square it: (-6)² = 36.
– So, we rewrite x² – 12x as: (x – 6)² – 36.
2. For the y terms: y² – 2y
– Take half of -2, which is -1, and square it: (-1)² = 1.
– So, we rewrite y² – 2y as: (y – 1)² – 1.
Substituting these back into our equation gives:
(x – 6)² – 36 + (y – 1)² – 1 = -12
Now combine and rearrange:
(x – 6)² + (y – 1)² = -12 + 36 + 1
(x – 6)² + (y – 1)² = 25
Now we have it in the standard form:
(x – 6)² + (y – 1)² = 5²
From this equation, we can see that the center (h, k) of the circle is (6, 1).
Answer: The center of the circle is (6, 1).