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}+6 x + 4 y + 9 = 0 )

the equation in standard form is ( (x + 3)^{2}+(y + 2)^{2}=4 ).
(simplify your answer.)

the center of the circle is ( (-3,-2) ).
(type an ordered pair)

the radius of the circle is ( r = 2 ).

use the graphing tool to graph the circle.

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graph

Explanation:

Step1: Group the x - terms and y - terms

$$x^{2}+6x+y^{2}+4y+9 = 0$$

Step2: Complete the square for the x - terms

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

Step3: Complete the square for the y - terms

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

Step4: Substitute back into the original equation

\((x + 3)^{2}-9+(y + 2)^{2}-4+9=0\)
Simplify the equation: \((x + 3)^{2}+(y + 2)^{2}=4\)

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.

Comparing \((x + 3)^{2}+(y + 2)^{2}=4\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-3,k=-2,r = 2\)

Answer:

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