Finding The Center And Radius Of A Circle Equation $x^2+y^2-6x+8y=0$

Introduction

In the realm of analytic geometry, circles hold a fundamental position. They are not just shapes we encounter daily; they are also mathematical objects described by precise equations. Understanding these equations allows us to extract crucial information about the circle, such as its center and radius. This article delves into the process of determining the center and radius of a circle given its equation in general form. We will focus on a specific example: the equation $x^2 + y^2 - 6x + 8y = 0$. By the end of this discussion, you will have a solid grasp of how to transform the general equation of a circle into its standard form, thereby revealing its center and radius. This skill is essential not only for academic pursuits but also for various practical applications in fields like engineering, computer graphics, and physics. So, let's embark on this geometric journey and unravel the secrets hidden within the equation of a circle.

The General and Standard Forms of a Circle's Equation

Before we tackle the specific problem, it's crucial to understand the two primary forms of a circle's equation: the general form and the standard form. The general form of a circle's equation is given by:

$x^2 + y^2 + Dx + Ey + F = 0$

where D, E, and F are constants. This form, while complete, doesn't immediately reveal the circle's center and radius. On the other hand, the standard form of a circle's equation is:

$(x - h)^2 + (y - k)^2 = r^2$

Here, $(h, k)$ represents the coordinates of the center of the circle, and $r$ denotes the radius. The standard form provides a clear and direct way to identify the circle's key properties. Our goal is to convert the given equation, which is in general form, into the standard form. This conversion involves a technique called "completing the square," a powerful algebraic tool that will allow us to rewrite the equation in a more revealing format. Understanding the difference between these two forms and the process of converting between them is fundamental to solving circle-related problems in analytic geometry.

Completing the Square: The Key to Unlocking the Circle's Secrets

The technique of completing the square is the cornerstone of converting the general form of a circle's equation into its standard form. This method allows us to rewrite quadratic expressions as perfect squares, which is exactly what we need to do to get the equation into the $(x - h)^2 + (y - k)^2 = r^2$ format. The process involves taking half of the coefficient of the x term, squaring it, and adding it to both sides of the equation. We repeat this process for the y term as well. This might sound a bit abstract, but it becomes clear with an example. Let's consider the x terms in our equation, $x^2 - 6x$. To complete the square, we take half of -6, which is -3, and square it, resulting in 9. Similarly, for the y terms, $y^2 + 8y$, we take half of 8, which is 4, and square it, resulting in 16. By adding these values strategically to both sides of the equation, we can create perfect square trinomials that can be easily factored. Mastering the technique of completing the square is not only crucial for solving circle problems but also for a wide range of algebraic manipulations and problem-solving in mathematics. It's a skill that truly unlocks the power of algebraic transformations.

Applying Completing the Square to the Equation $x^2 + y^2 - 6x + 8y = 0$

Now, let's apply the completing the square technique to the given equation: $x^2 + y^2 - 6x + 8y = 0$. The first step is to group the x terms and the y terms together: $(x^2 - 6x) + (y^2 + 8y) = 0$. Next, we focus on completing the square for the x terms. As we discussed earlier, half of -6 is -3, and squaring it gives us 9. So, we add 9 to both sides of the equation. Then, we move on to the y terms. Half of 8 is 4, and squaring it gives us 16. We add 16 to both sides of the equation as well. This gives us: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 0 + 9 + 16$. Now, we can factor the expressions in parentheses. The expression $x^2 - 6x + 9$ factors into $(x - 3)^2$, and the expression $y^2 + 8y + 16$ factors into $(y + 4)^2$. On the right side of the equation, we have $9 + 16 = 25$. So, our equation now looks like this: $(x - 3)^2 + (y + 4)^2 = 25$. This is the standard form of the circle's equation, and from here, we can easily identify the center and radius.

Identifying the Center and Radius from the Standard Form

Once we have the equation in the standard form, $(x - h)^2 + (y - k)^2 = r^2$, identifying the center and radius becomes straightforward. The center of the circle is given by the coordinates $(h, k)$, and the radius is the square root of $r^2$. In our case, the equation is $(x - 3)^2 + (y + 4)^2 = 25$. Comparing this with the standard form, we can see that $h = 3$, $k = -4$ (note that the equation has $(y + 4)$, which is equivalent to $(y - (-4))$), and $r^2 = 25$. Therefore, the center of the circle is $(3, -4)$, and the radius is the square root of 25, which is 5. It's important to pay close attention to the signs when identifying the center coordinates, as a common mistake is to misinterpret the signs in the equation. Understanding how to extract this information directly from the standard form is a key skill in analytic geometry and allows us to quickly analyze the properties of a circle given its equation. This ability to connect the algebraic representation of a circle with its geometric attributes highlights the power of analytic geometry.

The Solution and Conclusion

In conclusion, by applying the technique of completing the square to the equation $x^2 + y^2 - 6x + 8y = 0$, we successfully transformed it into the standard form: $(x - 3)^2 + (y + 4)^2 = 25$. From this standard form, we readily identified the center of the circle as $(3, -4)$ and the radius as 5 units. Therefore, the correct answer is A. $(3, -4), 5$ units. This exercise demonstrates the power of algebraic manipulation in revealing geometric properties. The ability to convert between the general and standard forms of a circle's equation is a valuable skill in mathematics and its applications. It allows us to quickly determine key characteristics of a circle, such as its center and radius, which are essential for various problem-solving scenarios. This process not only reinforces our understanding of circles but also enhances our algebraic proficiency. Mastering these techniques opens doors to exploring more advanced concepts in geometry and related fields. The journey through this problem highlights the elegance and interconnectedness of mathematical concepts.