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