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Completing the square is a crucial algebraic technique used to transform a quadratic expression into a perfect square trinomial. This method is particularly useful when working with equations of circles, as it helps to identify the circle's center and radius.
To complete the square for a term like \( x^2 + x \), follow these steps:

• Identify the coefficient of the \( x \) term, which is 1 in this case.
• Divide the coefficient by 2, giving you \( \frac{1}{2} \).
• Square this result to get \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).
• Add and subtract \( \frac{1}{4} \) within the equation to balance it, forming a complete square as \( (x + \frac{1}{2})^2 \).

Using completing the square simplifies the equation into a form that reveals more about the graph. In this problem, it transforms \( x^2 + x + y^2 = 0 \) into \( (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \), making the characteristics of the circle obvious.

In the standard equation of a circle, \( (x - h)^2 + (y - k)^2 = r^2 \), the terms \( h \) and \( k \) represent the center's coordinates \((h, k)\) of the circle. Knowing the center is essential to understanding the position and orientation of a circle in the coordinate plane.
For our specific example, after completing the square, we rewrite the equation as \( (x + \frac{1}{2})^2 + (y - 0)^2 = \frac{1}{4} \).

• The expression \( (x + \frac{1}{2})^2 \) indicates that \( h = -\frac{1}{2} \).
• The \( y^2 \) term simplifies directly to \( (y - 0)^2 \), showing that \( k = 0 \).

Thus, the center of the circle is located at \( (-\frac{1}{2}, 0) \). This tells us that the circle is shifted to the left by \( \frac{1}{2} \) from the origin, but aligns with the x-axis because the y-coordinate remains at 0. Recognizing the center helps in graphing the circle and understanding its geometric properties.

The radius is a fundamental characteristic of a circle. It represents the distance from the center of the circle to any point on its circumference. In the circle's equation form \( (x - h)^2 + (y - k)^2 = r^2 \), the radius is denoted by \( r \).
From our simplified equation \( (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \), it's clear that:

• The right side of the equation is \( \frac{1}{4} \).
• This value corresponds to \( r^2 \), so the radius \( r \) is \( \sqrt{\frac{1}{4}} \).
• Calculating further gives \( r = \frac{1}{2} \).

Therefore, the circle has a small radius of \( \frac{1}{2} \). This indicates that any point on the circle is exactly \( \frac{1}{2} \) units away from the center, \( (-\frac{1}{2}, 0) \). Understanding the radius is quite important not only in visualization but also in applications involving the circle, such as calculations of area or circumferences.