Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^{2}+y^{2}+6x - 4y - 12 = 0$
(complete your answer.)
the center of the circle is $(-4,2)$
(type an ordered pair.)
the radius of the circle is $r = 4\sqrt{2}$
use the graphing tool to graph the circle.
click to enlarge graph

Explanation:

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

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

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

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

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

Step4: Substitute back into the equation

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The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Answer:

The center of the circle is \((-3,2)\) and the radius \(r = 5\)