Question

x² + y² - 6x + 6y + 9 = 0 is the equation of a circle with center (h,k) and radius r for: h = and k = and r = graph the circle.

Explanation:

Step1: Rewrite the equation in standard form

The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). We need to complete the square for the given equation \(x^2 + y^2 - 6x + 6y + 9 = 0\).

Group the \(x\) terms and \(y\) terms:
\((x^2 - 6x) + (y^2 + 6y) = -9\)

Complete the square for \(x\): take half of \(-6\) (which is \(-3\)), square it (\(9\)), and add it to the \(x\) group.
Complete the square for \(y\): take half of \(6\) (which is \(3\)), square it (\(9\)), and add it to the \(y\) group.

So we have:
\((x^2 - 6x + 9) + (y^2 + 6y + 9) = -9 + 9 + 9\)

Step2: Simplify the equation

Simplify each group:
\((x - 3)^2 + (y + 3)^2 = 9\)

Now we can compare with the standard form \((x - h)^2 + (y - k)^2 = r^2\).

Step3: Identify \(h\), \(k\), and \(r\)

From \((x - 3)^2 + (y - (-3))^2 = 3^2\), we can see that:

• \(h = 3\) (because the term with \(x\) is \((x - 3)^2\))
• \(k = -3\) (because the term with \(y\) is \((y - (-3))^2\))
• \(r = 3\) (because \(r^2 = 9\), so \(r = \sqrt{9} = 3\))

Answer:

\(h = \boxed{3}\)
\(k = \boxed{-3}\)
\(r = \boxed{3}\)