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.

x² + y² + 6x + 6y + 14 = 0

the equation in standard form is
(simplify your answer.)

the center of the circle is
(type an ordered pair.)

the radius of the circle is r =

use the graphing tool to graph the circle.

click to enlarge graph

Explanation:

Step1: Rearrange the equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.
Given \(x^{2}+y^{2}+6x + 6y+14 = 0\).
Group \(x\) - terms and \(y\) - terms: \((x^{2}+6x)+(y^{2}+6y)=- 14\).

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

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

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

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

Step4: Substitute back into the equation

\((x + 3)^{2}-9+(y + 3)^{2}-9=-14\).
Simplify: \((x + 3)^{2}+(y + 3)^{2}-18=-14\).
Then \((x + 3)^{2}+(y + 3)^{2}=4\).

Answer:

The equation in standard form is \((x + 3)^{2}+(y + 3)^{2}=4\).
The center of the circle is \((-3,-3)\).
The radius of the circle is \(r = 2\).