Question

convert the following equation for a circle to standard form $(x - h)^{2}+(y - k)^{2}=r^{2}$.
$x^{2}+y^{2}+4x - 6y - 23 = 0$
what is the center of the circle? enter your answer as an ordered pair.
what is the radius of the circle?

Explanation:

Step1: Group \(x\) and \(y\) terms

$$
LATEXBLOCK0
$$

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

For \(x^{2}+4x\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = x\), \(2b=4\Rightarrow b = 2\). Then \(x^{2}+4x=(x + 2)^{2}-4\)

Step3: Complete the square for \(y\) terms

For \(y^{2}-6y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2b = 6\Rightarrow b=3\). Then \(y^{2}-6y=(y - 3)^{2}-9\)

Step4: Substitute back into the equation

$$
LATEXBLOCK1
$$

Answer:

• Center: \((-2,3)\)
• Radius: \(6\)