Analyzing Circle Equation X² + Y² + 4x - 6y - 36 = 0 True Statements And Standard Form Conversion

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Let's dive into the fascinating world of circles and explore the equation x² + y² + 4x - 6y - 36 = 0. This equation represents a circle in the Cartesian plane, and by carefully analyzing it, we can uncover its key properties, including its center and radius. In this article, we will embark on a step-by-step journey to transform the given equation into its standard form, which will reveal the circle's characteristics and allow us to answer various questions about it. Before starting with a detailed breakdown, let's mention the significance of understanding circle equations, especially in fields like geometry, calculus, and physics. The ability to manipulate and interpret these equations is fundamental in solving problems related to circular motion, conic sections, and various geometric constructions. We'll also dissect some common misconceptions and pitfalls that students often encounter while dealing with circle equations. So, whether you are a student grappling with circle geometry or a math enthusiast seeking a deeper understanding, this guide aims to provide you with a comprehensive and insightful exploration of the given equation.

The journey begins with recognizing the general form of a circle's equation and understanding how to convert it into the standard form. The general form, which is what we have here, x² + y² + 4x - 6y - 36 = 0, isn't immediately helpful for identifying the circle's center and radius. That's where the standard form comes in handy. The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle and r represents its radius. Our mission is to transform the given general form into this standard form, and to do so, we'll employ a technique called completing the square. Completing the square is a powerful algebraic method that allows us to rewrite quadratic expressions in a more convenient form. In the context of circle equations, it helps us group the x terms and y terms separately and rewrite them as perfect squares. This will lead us directly to the standard form, revealing the circle's center and radius. We will go through each step meticulously, explaining the logic behind each operation and highlighting potential areas where mistakes can occur. By the end of this process, you'll not only be able to solve this particular problem but also be well-equipped to handle a wide range of circle equation problems.

Step-by-Step Conversion to Standard Form

The key to unlocking the secrets of this circle lies in transforming the equation into its standard form. The standard form of a circle's equation, as we discussed earlier, is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. To achieve this transformation, we will employ the method of completing the square. This method involves manipulating the equation algebraically to create perfect square trinomials for both the x and y terms. Let's break down the process step by step.

  1. Rearrange and Group Terms: The first crucial step is to rearrange the given equation and group the x terms and y terms together. This sets the stage for completing the square separately for each variable. We start with the equation x² + y² + 4x - 6y - 36 = 0. By rearranging the terms, we get (x² + 4x) + (y² - 6y) = 36. Notice that we've moved the constant term, -36, to the right side of the equation, as this will be essential for determining the radius later on. The grouping of x and y terms allows us to focus on each variable individually, making the process of completing the square more manageable. This step might seem simple, but it's a critical foundation for the subsequent steps. A clear and organized arrangement of terms is paramount for avoiding errors and ensuring a smooth transition to the next stage.

  2. Completing the Square for x: Now comes the heart of the transformation: completing the square for the x terms. To complete the square for the expression x² + 4x, we need to add a constant term that will make it a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. The constant term we need to add is determined by taking half of the coefficient of the x term (which is 4), squaring it, and adding the result to both sides of the equation. Half of 4 is 2, and 2 squared is 4. So, we add 4 to both sides of the equation: (x² + 4x + 4) + (y² - 6y) = 36 + 4. Now, the expression x² + 4x + 4 is a perfect square trinomial, which can be factored as (x + 2)². This transformation is a key step in converting the equation to standard form. By completing the square, we've rewritten the x terms in a form that directly relates to the x coordinate of the circle's center.

  3. Completing the Square for y: We repeat the same process for the y terms. To complete the square for the expression y² - 6y, we need to add a constant term that will make it a perfect square trinomial. The coefficient of the y term is -6. Half of -6 is -3, and -3 squared is 9. So, we add 9 to both sides of the equation: (x² + 4x + 4) + (y² - 6y + 9) = 36 + 4 + 9. Now, the expression y² - 6y + 9 is a perfect square trinomial, which can be factored as (y - 3)². This step mirrors the process we used for the x terms, and it's equally crucial for obtaining the standard form of the equation. By completing the square for the y terms, we've rewritten them in a form that directly relates to the y coordinate of the circle's center.

  4. Rewrite in Standard Form: Now that we've completed the square for both x and y terms, we can rewrite the equation in standard form. The left side of the equation now consists of two perfect square trinomials, and the right side is a constant. Substituting the factored forms, we get (x + 2)² + (y - 3)² = 49. This is the standard form of the circle's equation! From this form, we can directly identify the center and the radius of the circle. The equation is now in the form (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. Comparing our equation to the standard form, we can see that h = -2, k = 3, and r² = 49. Therefore, the center of the circle is (-2, 3), and the radius is √49 = 7. This is a significant achievement, as we've successfully transformed the original equation into a form that directly reveals the circle's key characteristics.

Identifying True Statements

With the equation now in standard form, we can confidently evaluate the given statements and determine which ones are true. The standard form, (x + 2)² + (y - 3)² = 49, provides us with all the information we need to analyze the statements.

Let's revisit the statements:

  • A. To begin converting the equation to standard form, subtract 36 from both sides.

    This statement is incorrect. As we saw in the first step of our conversion process, we actually added 36 to both sides of the equation to isolate the constant term. Subtracting 36 would move the constant term further away from the variables, making it harder to complete the square. The correct initial step is to add 36 to both sides, which gives us (x² + 4x) + (y² - 6y) = 36, setting the stage for completing the square.

  • B. To complete the square for the x terms, add 4 to both sides.

    This statement is correct. As we discussed in detail, completing the square for the x terms, x² + 4x, requires adding the square of half the coefficient of the x term. Half of 4 is 2, and 2 squared is 4. Adding 4 to both sides allows us to rewrite the x terms as a perfect square trinomial, (x + 2)². This is a crucial step in the conversion process, as it helps us move closer to the standard form of the equation. The ability to correctly identify the constant term needed to complete the square is a fundamental skill in algebra and is essential for working with quadratic equations and conic sections.

Conclusion

In conclusion, by meticulously applying the method of completing the square, we successfully transformed the equation x² + y² + 4x - 6y - 36 = 0 into its standard form, (x + 2)² + (y - 3)² = 49. This allowed us to identify the center of the circle as (-2, 3) and the radius as 7. We also analyzed the given statements, confirming that statement B is true, while statement A is false. Understanding how to manipulate circle equations and convert them into standard form is a valuable skill in mathematics, with applications in various fields. This exploration not only provides a solution to the specific problem but also equips you with the knowledge and skills to tackle similar challenges in the future. The process of completing the square, in particular, is a versatile technique that extends beyond circle equations and is applicable to a wide range of algebraic problems. Remember, the key to mastering these concepts is practice and a thorough understanding of the underlying principles.