Question

the general form of a circle is $x^{2}+y^{2}-6x + 4y+9 = 0$.
complete the square to write the equation in standard form and find the center and radius of the circle.
use the keypad to enter your answers in the boxes.
the equation of the circle in standard form is $(xsquare)^{2}+(ysquare)^{2}=square$.
the center of the circle is at $square$.
the radius of the circle is $square$.

Explanation:

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

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

For \(x^{2}-6x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2b=6\Rightarrow b = 3\). Then \(x^{2}-6x=(x - 3)^{2}-9\)

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

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

Step4: Substitute back into the equation

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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.

Answer:

The equation of the circle in standard form is \((x - 3)^{2}+(y+ 2)^{2}=4\)
The center of the circle is at \((3,-2)\)
The radius of the circle is \(2\)