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Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This is particularly useful when working with the equation of a circle since it helps us rewrite the equation in a form that reveals the circle's center and radius.

In the given problem, we start with the equation: \(x^{2}+y^{2}-6 y=0\).
To complete the square for the y terms, follow these steps:

• Identify the coefficient of y. Here, it is -6.
• Halve this coefficient to get -3.
• Square this result to get 9.

Then, we add and subtract 9 within the equation, ensuring the equation remains balanced: \(x^{2}+(y^{2}-6y+9-9)=0\). This simplifies to \(x^{2}+(y-3)^{2}-9=0\).

Completing the square helps us convert the quadratic equation into a recognizable form that makes it easy to identify the circle's properties.

The center of a circle in the standard form equation \((x-h)^{2}+(y-k)^{2}=r^{2}\) is given by the coordinates \((h, k)\).

After completing the square, the equation from our exercise is now: \((x-0)^{2}+(y-3)^{2}=9\).
Here, it is clear that \((h, k)\) are the coordinates of the center. Comparing the equation to the standard form, we see

• h = 0
• k = 3

The coordinates of the center of the circle are therefore \((0, 3)\).

Finding the center is crucial because it helps us understand the position of the circle within the coordinate plane and assists with graphing.

The radius of a circle is the distance from the center to any point on the circle. In the standard form equation \((x-h)^{2}+(y-k)^{2}=r^{2}\), the radius \(r\) is determined from the term \(r^{2}\).

In our given problem, once the completing the square method was applied, the equation became: \((x-0)^{2}+(y-3)^{2}=9\). Here, \(r^{2}=9\).
To find the radius, take the square root of this term:

• \(r = \sqrt{9} = 3\)

The radius of the circle is therefore 3 units.

Knowing the radius helps in understanding the size of the circle and is especially useful when constructing or analyzing the circle's geometry.