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Completing the square is a fundamental algebraic technique used to simplify quadratic expressions. It helps transform a quadratic expression into a perfect square trinomial, making it easier to solve or graph equations like circles. For the equation given in the original exercise, completing the square was needed for both the \(x\)- and \(y\)-terms in order to rewrite the circle's equation in standard form.
To complete the square for any term \(ax^2 + bx\):

• Take half of the coefficient of \(x\), which is \(b/2\).
• Square this value to get \((b/2)^2\).
• Add and subtract this square inside the equation to maintain equality.

For instance, in the equation \(x^2 - 2x\), we took half of \(-2\) to get \(-1\), then squared it to get 1. We added 1 and subtracted 1 to yield \((x-1)^2 - 1\).Another example is for the \(y\)-terms \(y^2 + 4y\), where we took half of 4, squared it to get 4, and rearranged as \((y+2)^2 - 4\).
This step by step completes the square process brings the quadratic expression into a form that’s ready to be rearranged into a circle's standard equation.

The standard form of a circle's equation is a powerful representation that simplifies identifying the circle's properties. It is expressed as \((x - h)^2 + (y - k)^2 = r^2\). In this format, \((h, k)\) identifies the circle’s center, and \(r\) is the radius.
Once an equation is rearranged to fit this form, as done in the exercise, you can readily identify the circle’s main features without further calculation.
The transformation of the original equation by completing the square allowed it to be rewritten in the needed standard form, \((x-1)^2 + (y+2)^2 = 5\). This equation illustrates a circle centered at \((1, -2)\) with a radius of \(\sqrt{5}\).
Converting equations to this standard form helps quickly discern whether an equation represents a circle and facilitates the process of graphically plotting the circle by providing clear coordinates for its center and a specific measure for its radius.

Identifying the center and radius of a circle is made easy once the equation is in its standard form. The center of the circle, as indicated in the format \((x - h)^2 + (y - k)^2 = r^2\), is \((h, k)\), and the radius \(r\) is found by taking the square root of \(r^2\).
For the example \((x-1)^2 + (y+2)^2 = 5\), we can see:

• The center \((h, k)\) is \((1, -2)\), derived directly from \((x-1)\) and \((y+2)\).
• The radius \(r\) is calculated by taking the square root of 5, resulting in \(\sqrt{5}\).

Understanding this determination of center and radius is crucial. It enables visualization of the circle in a 2D plane, as the center shows where the circle is "anchored," and the radius tells you how "big" the circle is. These forms allow you to effectively transition from equation algebra to graphical representation.