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.
$x^{2}+y^{2}+6x + 6y+14 = 0$
the equation in standard form is
(simplify your answer.)

Explanation:

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

Group the \(x -\)terms and \(y -\)terms together:
\((x^{2}+6x)+(y^{2}+6y)+14 = 0\)

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

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

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

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

Step4: Substitute back into the equation

Substitute \(x^{2}+6x=(x + 3)^{2}-9\) and \(y^{2}+6y=(y + 3)^{2}-9\) into \((x^{2}+6x)+(y^{2}+6y)+14 = 0\)
\((x + 3)^{2}-9+(y + 3)^{2}-9+14=0\)

Step5: Simplify the equation

Simplify the left - hand side:
\((x + 3)^{2}+(y + 3)^{2}-18 + 14=0\)
\((x + 3)^{2}+(y + 3)^{2}-4=0\)
\((x + 3)^{2}+(y + 3)^{2}=4\)

Answer:

The equation in standard form is \((x + 3)^{2}+(y + 3)^{2}=4\)