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.
$x^{2}+y^{2}+8x - 2y - 19 = 0$
the equation of the circle in standard form is
$(x + 4)^{2}+(y - 1)^{2}=36$
(simplify your answer.)
the center of the circle is $(-4,1)$
(type an ordered pair.)
the radius of the circle is $r = 6$
use the graphing tool to graph the circle.

Explanation:

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

$$
LATEXBLOCK0
$$

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

For \(x^{2}+8x\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 8x\Rightarrow b = 4\), so \(x^{2}+8x=(x + 4)^{2}-16\)

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

For \(y^{2}-2y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=y\), \(2ab=2y\Rightarrow b = 1\), so \(y^{2}-2y=(y - 1)^{2}-1\)

Step4: Substitute back into the equation

$$
LATEXBLOCK1
$$

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 \((x + 4)^{2}+(y - 1)^{2}=36=(x-(-4))^{2}+(y - 1)^{2}=6^{2}\)

Answer:

The equation of the circle in standard form is \((x + 4)^{2}+(y - 1)^{2}=36\). The center of the circle is \((-4,1)\). The radius of the circle is \(r = 6\)