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Completing the square is a technique used to transform quadratic equations into a format that is easier to work with. This method plays a crucial role in many areas, including solving equations of circles as seen in the given problem. In the context of the equation for a circle, completing the square helps in converting the general form of a quadratic equation into the standard form of a circle equation.

The general form of a circle equation is given by the sum of squares of the variables, plus additional linear terms, such as:

• \((x^2 + y^2 + Dx + Ey = F)\)

To complete the square for each variable:

• Take the linear terms, e.g., \(Dx=8x\), and divide the coefficient of the variable by two, then square it.
• Add and subtract this squared value within the equation.
• Reformat the equation into perfect square trinomials.
• For \(x^2 + 8x\), this results in \( (x + 4)^2\).

Similarly, complete the square for the \(y\) terms. Once all terms are completed, the equation can be refactored into the standard form, \((x - h)^2 + (y - k)^2 = r^2\). This form makes it easier to identify the circle's center and radius.

Understanding how to find the intersection points of a circle with an axis or another line is vital in geometry. In this exercise, we're focused on the intersection with the \(y\)-axis. When seeking intersection points where a circle meets the \(y\)-axis, we substitute \(x = 0\) in the circle's equation. This is because any point on the \(y\)-axis has an \(x\)-coordinate of 0.

For example, in the equation \((x + 4)^2 + (y - 3)^2 = 1\), substituting \(x = 0\) transforms the equation into:

• \(16 + (y - 3)^2 = 1\)

However, solving leads to \((y - 3)^2 = -15\), which implies no real solutions exist. This conclusion means the circle does not intersect the \(y\)-axis at any point. This highlights that if simplifying reveals a negative on the right side (of the \( (y - 3)^2 = ext{value} \) equation), there cannot be real solutions—emphasizing a key aspect of circle intersections.

Finding the center and radius of a circle from its equation gives us valuable geometric information. Transforming the equation into the standard form \((x - h)^2 + (y - k)^2 = r^2\) reveals these properties clearly.

Using the result of completing the square, the given equation becomes \((x + 4)^2 + (y - 3)^2 = 1\). This is directly comparable to the standard form, where:

• \(h = -4\), indicating the center's \((x)\)-coordinate.
• \(k = 3\), the \((y)\)-coordinate of the center.
• The radius \(r\) is the square root of the right-hand side: \(r = \sqrt{1} = 1\).

Thus, the center of this circle is at \((-4, 3)\) and its radius is \1\. Understanding these transformations from algebra to geometry is crucial, as it allows us to visually interpret and place the circle accurately on a coordinate plane.