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.
$x^{2}+y^{2}+4x - 2y - 31 = 0$
the equation of the circle in standard form is
(simplify your answer)

Explanation:

Step1: Group x and y terms

$$(x^{2}+4x)+(y^{2}-2y)=31$$

Step2: Complete the square for x - terms

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

Step3: Complete the square for y - terms

For \(y^{2}-2y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\) and \(2ab = 2y\Rightarrow b=1\). Then \(y^{2}-2y=(y - 1)^{2}-1\)

Step4: Substitute back into the equation

\((x + 2)^{2}-4+(y - 1)^{2}-1=31\)
\((x + 2)^{2}+(y - 1)^{2}=31 + 4+1\)

Answer:

The equation of the circle in standard form is \((x + 2)^{2}+(y - 1)^{2}=36\)