Question

consider the following equation of a circle.
x^{2}+y^{2}-6x + 6y+9 = 0
step 2 of 3: find the radius, r.
answer
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Explanation:

Step1: Rewrite the equation in standard form

Complete the square for \(x\) and \(y\) terms.
The general equation of a circle is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center and \(r\) is the radius.
For the \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\).
For the \(y\) - terms: \(y^{2}+6y=(y + 3)^{2}-9\).
The given 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

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

Step3: Identify the radius

Comparing \((x - 3)^{2}+(y + 3)^{2}=9\) with \((x - a)^2+(y - b)^2=r^2\), we have \(r^{2}=9\), so \(r = 3\).

Answer:

3