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.
$x^{2}+y^{2}+6x + 4y+9 = 0$
the equation in standard form is
(simplify your answer.)

Explanation:

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

Group the \(x\) - terms and \(y\) - terms together:
\((x^{2}+6x)+(y^{2}+4y)+9 = 0\)

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

For the \(x\) - terms \(x^{2}+6x\), use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here \(a=x\) and \(2ab = 6x\), so \(b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\).
For the \(y\) - terms \(y^{2}+4y\), use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here \(a = y\) and \(2ab=4y\), so \(b = 2\). Then \(y^{2}+4y=(y + 2)^{2}-4\).
Substitute these into the equation:
\((x + 3)^{2}-9+(y + 2)^{2}-4+9 = 0\)

Step3: Simplify the equation

Simplify the left - hand side of the equation:
\((x + 3)^{2}+(y + 2)^{2}-4=0\)
Add \(4\) to both sides of the equation:
\((x + 3)^{2}+(y + 2)^{2}=4\)

Answer:

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