Question

the equation of a circle in expanded form is shown. what is the equation of the circle in standard form? use the drop - down menus to complete the equation. $x^{2}+y^{2}-8x + 4y-16 = 0$ click the arrows to choose an answer from each menu. $(\text{choose...})^{2}+(\text{choose...})^{2}=\text{choose...}^{2}$

Explanation:

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

$$(x^{2}-8x)+(y^{2}+4y)=16$$

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

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

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

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

Step4: Substitute back into the equation

\((x - 4)^{2}-16+(y + 2)^{2}-4=16\)
\((x - 4)^{2}+(y + 2)^{2}=16 + 16+4\)
\((x - 4)^{2}+(y + 2)^{2}=36=6^{2}\)

Answer:

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