Decoding The Circle Equation $x^2 + Y^2 - 2x + 6y + 9 = 0$ Center, Radius, And Graph

Introduction

In the fascinating world of analytic geometry, circles hold a special place. Their elegant symmetry and fundamental properties make them a cornerstone of mathematical understanding. A circle, defined as the set of all points equidistant from a central point, can be represented by a concise algebraic equation. In this article, we embark on a journey to dissect the equation $x^2 + y^2 - 2x + 6y + 9 = 0$, revealing its hidden secrets and geometric essence. We will navigate the process of transforming this equation into the standard form of a circle's equation, allowing us to pinpoint the circle's center coordinates, calculate its radius, and ultimately, visualize its graceful form on a graph.

This comprehensive exploration is designed to provide a clear and accessible understanding of circle equations, empowering you to analyze and interpret these equations with confidence. Whether you're a student delving into the intricacies of geometry or a mathematics enthusiast seeking to expand your knowledge, this article offers a step-by-step guide to unraveling the mysteries of circles and their algebraic representations.

Transforming the Equation: Completing the Square

Our mission begins with the equation $x^2 + y^2 - 2x + 6y + 9 = 0$. This equation, while representing a circle, isn't in its most revealing form. To unlock the circle's properties, we must transform it into the standard equation of a circle, which is $(x - h)^2 + (y - k)^2 = r^2$. Here, $(h, k)$ represents the coordinates of the circle's center, and $r$ denotes its radius. The key to this transformation lies in a powerful algebraic technique known as "completing the square."

Completing the square is a method used to rewrite quadratic expressions into a perfect square trinomial, which can then be factored into a squared binomial. This technique is crucial for converting the given equation into the standard circle equation. Our first step is to group the $x$ terms and the $y$ terms together, preparing them for the completing the square process. We rewrite the equation as $(x^2 - 2x) + (y^2 + 6y) + 9 = 0$. Notice that we've strategically grouped the terms to isolate the quadratic and linear components of both $x$ and $y$.

Now, let's focus on the $x$ terms, $x^2 - 2x$. To complete the square, we need to add a constant term that will make this expression a perfect square trinomial. This constant is determined by taking half of the coefficient of the $x$ term (-2), squaring it, and adding the result. Half of -2 is -1, and squaring -1 gives us 1. So, we add 1 to the expression. Simultaneously, we apply the same logic to the $y$ terms, $y^2 + 6y$. Half of 6 is 3, and squaring 3 yields 9. We add 9 to the $y$ expression. However, to maintain the equation's balance, we must add these same constants to the right side of the equation as well. Therefore, we subtract 1 and 9 to the constant 9 on the left side of the equation. This leads us to: $(x^2 - 2x + 1) + (y^2 + 6y + 9) + 9 - 1 - 9 = 0$.

The expressions within the parentheses are now perfect square trinomials. We can factor them into squared binomials: $(x - 1)^2 + (y + 3)^2 - 1 = 0$. Moving the constant term to the right side of the equation gives us the standard form: $(x - 1)^2 + (y + 3)^2 = 1$. This is the equation of our circle in its most informative form.

Decoding the Circle's Properties: Center and Radius

With the equation now in the standard form, $(x - 1)^2 + (y + 3)^2 = 1$, the circle's properties are revealed with remarkable clarity. By comparing this equation to the general standard form, $(x - h)^2 + (y - k)^2 = r^2$, we can directly identify the center coordinates $(h, k)$ and the radius $r$.

In our equation, we see that $h = 1$ and $k = -3$ (note the sign change due to the subtraction in the standard form). Therefore, the center of the circle is located at the point $(1, -3)$. This point serves as the anchor around which the entire circle is drawn, equidistant from every point on the circumference.

Next, we turn our attention to the right side of the equation, which represents $r^2$. In our case, $r^2 = 1$. To find the radius $r$, we take the square root of both sides: $r = \_1 = 1$. Thus, the radius of the circle is 1 unit. The radius dictates the circle's size, defining the distance from the center to any point on the circle's edge.

In summary, by transforming the original equation and comparing it to the standard form, we have successfully determined the circle's center to be $(1, -3)$ and its radius to be 1 unit. These two pieces of information are sufficient to fully describe and visualize the circle.

Graphing the Circle: A Visual Representation

Having determined the center and radius of the circle, we now embark on the final step: visualizing the circle through a graph. Graphing the circle provides a powerful way to understand its spatial representation and confirm our algebraic findings. To graph the circle, we will utilize the center coordinates $(1, -3)$ and the radius of 1 unit.

First, we establish a coordinate plane, with the horizontal axis representing the x-coordinates and the vertical axis representing the y-coordinates. We then locate the center of the circle at the point $(1, -3)$ and mark it on the plane. This point serves as the focal point for our circular construction.

Next, we use the radius to determine the circle's extent in all directions. Since the radius is 1 unit, we know that the circle extends 1 unit to the right, 1 unit to the left, 1 unit upwards, and 1 unit downwards from the center. We can mark these four points on the coordinate plane: $(2, -3)$, $(0, -3)$, $(1, -2)$, and $(1, -4)$. These points represent the circle's farthest reach in each cardinal direction.

Now, with the center and four key points established, we can sketch the circle. Imagine a smooth curve that passes through the four points we marked, maintaining a consistent distance of 1 unit from the center. This curve represents the circumference of the circle, and the resulting shape is the visual representation of the equation $x^2 + y^2 - 2x + 6y + 9 = 0$. The graph allows us to see the circle's position on the coordinate plane, its size determined by the radius, and its overall shape.

The graphical representation serves as a valuable confirmation of our algebraic calculations. We can visually verify that the circle is indeed centered at $(1, -3)$ and has a radius of 1 unit. This connection between the algebraic equation and the geometric representation underscores the beauty and power of analytic geometry.

Conclusion

In this exploration, we have successfully navigated the journey from an unfamiliar equation to a clear understanding of a circle's properties. We began with the equation $x^2 + y^2 - 2x + 6y + 9 = 0$ and, through the powerful technique of completing the square, transformed it into the standard form $(x - 1)^2 + (y + 3)^2 = 1$. This transformation allowed us to readily identify the circle's center coordinates as $(1, -3)$ and its radius as 1 unit.

Furthermore, we translated these algebraic insights into a visual representation by graphing the circle on a coordinate plane. The graph served as a tangible confirmation of our calculations, showcasing the circle's position, size, and shape. This process highlights the interconnectedness of algebra and geometry, demonstrating how equations can be used to describe and visualize geometric objects.

The ability to decode circle equations is a fundamental skill in mathematics, with applications ranging from geometry and calculus to physics and engineering. By mastering the techniques presented in this article, you are well-equipped to analyze and interpret circle equations, unlocking a deeper understanding of these fundamental geometric shapes. This journey into the world of circles has not only revealed the properties of a specific equation but has also illuminated the broader principles of analytic geometry, empowering you to explore the fascinating relationships between algebra and geometry.