Question

question 10
$x^{2}+y^{2}-2x + 6y + 6 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$ for:
$h=$
and
$k=$
and
$r=$
graph the circle.
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Explanation:

Step1: Complete the square for \(x\) terms

Group \(x\) terms: \(x^{2}-2x=(x - 1)^{2}-1\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 1\)).

Step2: Complete the square for \(y\) terms

Group \(y\) terms: \(y^{2}+6y=(y + 3)^{2}-9\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 3\)).

Step3: Rewrite the circle equation

The original equation \(x^{2}+y^{2}-2x + 6y+6 = 0\) becomes \((x - 1)^{2}-1+(y + 3)^{2}-9+6=0\).
Simplify to \((x - 1)^{2}+(y + 3)^{2}=4\).

Answer:

\(h = 1\)
\(k=-3\)
\(r = 2\)