Question

consider the following equation of a circle.
$x^{2}+y^{2}-6x + 6y+9 = 0$
step 1 of 3: find the center, $(h,k)$.
answer
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Explanation:

Step1: Rewrite the given equation

We complete the square for the \(x\) - terms and \(y\) - terms in the equation \(x^{2}+y^{2}-6x + 6y+9 = 0\).
The general form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle.
For the \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\) (using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\) and \(b = 3\)).
For the \(y\) - terms: \(y^{2}+6y=(y + 3)^{2}-9\) (using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=y\) and \(b = 3\)).
So the equation \(x^{2}+y^{2}-6x + 6y+9 = 0\) can be rewritten as \((x - 3)^{2}-9+(y + 3)^{2}-9+9 = 0\).

Step2: Simplify the equation

Simplify \((x - 3)^{2}-9+(y + 3)^{2}-9+9 = 0\) to \((x - 3)^{2}+(y + 3)^{2}=9\).

Answer:

\((3,-3)\)