Question

$x^{2}+y^{2}-6x + 8y + 9 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$ for:
$h = \square$
and
$k = \square$
and
$r = \square$
graph the circle.
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Explanation:

Step1: Rewrite the circle equation

The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). Given the equation \(x^2 + y^2 - 6x + 8y + 9 = 0\), we complete the square for \(x\) and \(y\) terms.
For \(x\)-terms: \(x^2 - 6x=(x - 3)^2 - 9\) (since \((x - a)^2=x^2 - 2ax+a^2\), here \(2a = 6\), so \(a = 3\)).
For \(y\)-terms: \(y^2 + 8y=(y + 4)^2 - 16\) (here \(2a=8\), so \(a = 4\)).
Substitute these into the original equation:

$$
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$$

Step2: Identify \(h\), \(k\), and \(r\)

Comparing \((x - 3)^2+(y + 4)^2 = 16\) with \((x - h)^2+(y - k)^2 = r^2\), we have:

• \(h = 3\) (since the term with \(x\) is \((x - 3)^2\))
• \(k=- 4\) (since the term with \(y\) is \((y - (-4))^2=(y + 4)^2\))
• \(r^2 = 16\), so \(r = 4\) (taking the positive square root as radius is non - negative)

Answer:

\(h = 3\), \(k=-4\), \(r = 4\)