Question

what is the center of a circle whose equation is $x^{2}+y^{2}+4x - 8y + 11 = 0$?
$(-2,4)$
$(-4,8)$
$(2,-4)$
$(4,-8)$

Explanation:

Step1: Complete the square for \(x\) terms

Group \(x\) terms: \(x^{2}+4x=(x + 2)^{2}-4\)

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

Group \(y\) terms: \(y^{2}-8y=(y - 4)^{2}-16\)

Step3: Rewrite the circle equation

Substitute into original equation: \((x + 2)^{2}-4+(y - 4)^{2}-16 + 11=0\)
Simplify: \((x + 2)^{2}+(y - 4)^{2}=9\)
The standard form of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center.

Answer:

A. \((-2,4)\)