Question

directions: type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
the equation of a circle is given by the following function.
$x^{2}+y^{2}-6x + 4y + 4 = 0$
the center of the circle is located at (□,□), and the diameter of the circle is □ units.

Explanation:

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

The general equation of a circle is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center and \(r\) is the radius.
Given \(x^{2}+y^{2}-6x + 4y+4 = 0\).
For the \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\) (using \((m - n)^2=m^{2}-2mn + n^{2}\), here \(m = x\), \(n = 3\), \(x^{2}-6x=x^{2}-2\times3x\), then add and subtract \(3^{2}\)).
For the \(y\) - terms: \(y^{2}+4y=(y + 2)^{2}-4\) (using \((m + n)^2=m^{2}+2mn + n^{2}\), here \(m = y\), \(n = 2\), \(y^{2}+4y=y^{2}+2\times2y\), then add and subtract \(2^{2}\)).
Substitute these into the original equation:
\((x - 3)^{2}-9+(y + 2)^{2}-4 + 4=0\).

Step2: Simplify the equation

Simplify \((x - 3)^{2}-9+(y + 2)^{2}-4 + 4=0\) to \((x - 3)^{2}+(y + 2)^{2}=9\).

Step3: Identify the center and radius

Comparing with \((x - a)^2+(y - b)^2=r^2\), the center \((a,b)=(3,-2)\) and \(r = 3\) (since \(r^{2}=9\)). The diameter \(d = 2r\).

Answer:

The center of the circle is at \(3,-2\), and the diameter of the circle is \(6\) units.