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Completing the square is a mathematical method used to simplify equations, making them easier to analyze or graph. When working with a circle's equation, this technique helps to rearrange the equation to a recognizable, standard form. This form is \((x-h)^2 + (y-k)^2 = r^2\), which clearly displays the circle's center and radius.

In our exercise with the equation \(x^2 + y^2 + 4x - 4y + 4 = 0\), we complete the square for both \(x\) and \(y\).

• For the \(x\) terms: Look at the coefficient of \(x\), which is 4. Divide it by 2 to get 2. Square this result to get 4.
• Insert \(x^2 + 4x + 4\) into the equation.
• For the \(y\) terms: Take the coefficient of \(y\), which is -4. Divide by 2 to get -2, square it to get 4.
• Insert \(y^2 - 4y + 4\) into the equation.

This process ingeniously rearranges terms into perfect square trinomials like \((x+2)^2\) and \((y-2)^2\). This step transforms the given circle equation into something more user-friendly, indicating how to obtain the circle's center and radius.

The center and radius of a circle provide vital information about its location and size on a graph. Once we complete the square for both \(x\) and \(y\), the equation takes the standard form. This form is \((x-h)^2 + (y-k)^2 = r^2\).

• The values \((h, k)\) show us the center of the circle. It's the point from which the circle reaches out equally in all directions.
• The value \(r\) shows the circle's radius. This is the distance from the center to any point on the circle's edge.

For our problem, after completing the square, we have the equation \((x+2)^2 + (y-2)^2 = 8\). This tells us that:

• The center of the circle is at the point \((-2, 2)\).
• The radius, derived from \(\sqrt{8}\), is \(2\sqrt{2}\).

Understanding these components is crucial when graphing the circle or identifying key characteristics like intercepts.

X-intercepts are the points where a circle crosses the x-axis. To find these intercepts, set \(y = 0\) in the circle's equation because the x-intercept occurs where the y-value is zero.

Using our example equation \((x+2)^2 + (y-2)^2 = 8\), and setting \(y = 0\), we have:

• \((x+2)^2 + (0-2)^2 = 8\)
• Which simplifies to \((x+2)^2 = 4\)
• This results in \(x = -2 \pm 2\)

Thus, the x-intercepts are at the points \((0, 0)\) and \((-4, 0)\).
Remember that intercepts give us exact points on the graph where the circle intersects the axes, providing a clear view of its position.

Y-intercepts are the points where a circle meets the y-axis. To locate these, we set \(x = 0\) in the circle's equation, as y-intercepts occur where the x-value is zero.

Starting from our simplified standard form equation \((x+2)^2 + (y-2)^2 = 8\), with \(x = 0\):

• Insert \(x = 0\) to get \((0+2)^2 + (y-2)^2 = 8\).
• This simplifies to \((y-2)^2 = 4\).
• The solutions are \(y = 2 \pm 2\).

We conclude that the y-intercepts are at points \((0, 0)\) and \((0, 4)\).
These intercepts reveal the precise places the circle makes contact with the y-axis, giving another way to verify the circle's plot on a coordinate grid.