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 } + 8 x - 4 y - 12 = 0 )

the equation of the circle 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.

Explanation:

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

Group \(x\) - terms \((x^{2}+8x)\) and \(y\) - terms \((y^{2}-4y)\) together:
\((x^{2}+8x)+(y^{2}-4y)=12\)

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

For \(x^{2}+8x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = x\) and \(2b=8\) (so \(b = 4\)). Then \(x^{2}+8x=(x + 4)^{2}-16\)

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

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

Step4: Substitute back into the equation

Substitute \((x + 4)^{2}-16+(y - 2)^{2}-4=12\)
\((x + 4)^{2}+(y - 2)^{2}=12 + 16+4\)
\((x + 4)^{2}+(y - 2)^{2}=32\)

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 \((x + 4)^{2}+(y - 2)^{2}=32\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-4,k = 2,r=\sqrt{32}=4\sqrt{2}\)

Answer:

The equation of the circle in standard form is \((x + 4)^{2}+(y - 2)^{2}=32\)
The center of the circle is \((-4,2)\)
The radius of the circle is \(r = 4\sqrt{2}\)