Finding The Center Of A Circle X² + Y² + 4x - 8y + 11 = 0

Finding the center of a circle given its equation is a fundamental concept in geometry, often encountered in various mathematical contexts and real-world applications. This article provides a detailed, step-by-step guide on how to determine the center of a circle when its equation is presented in the general form. We will specifically address the equation x² + y² + 4x - 8y + 11 = 0, walking you through the process of transforming it into the standard form, which readily reveals the circle's center coordinates. By mastering this technique, you'll gain a valuable skill applicable to a wide range of geometric problems.

Understanding the Standard Form of a Circle Equation

Before we delve into the specifics of solving the given equation, it's crucial to understand the standard form of a circle equation. The standard form is expressed as:

(x - h)² + (y - k)² = r²

where:

• (h, k) represents the coordinates of the center of the circle.
• r represents the radius of the circle.

This form is incredibly useful because it directly provides the center and radius, making it easy to visualize and analyze the circle's properties. Our goal is to manipulate the given equation, x² + y² + 4x - 8y + 11 = 0, into this standard form. This involves a technique called "completing the square," which we will explore in detail.

The Power of Completing the Square: Transforming the Equation

Completing the square is an algebraic method used to rewrite a quadratic expression into a perfect square trinomial plus a constant. This technique is essential for transforming the general form of a circle equation into the standard form. Let's apply this method to our equation, x² + y² + 4x - 8y + 11 = 0:

1. Group the x and y terms:

   (x² + 4x) + (y² - 8y) = -11

   We first group the terms containing 'x' and the terms containing 'y' together, moving the constant term to the right side of the equation. This sets the stage for completing the square for both the x and y components.

2. Complete the square for the x terms:

   To complete the square for the expression x² + 4x, we need to add and subtract the square of half the coefficient of the x term. The coefficient of the x term is 4, so half of it is 2, and the square of 2 is 4. Therefore, we add and subtract 4 within the parentheses:

   (x² + 4x + 4 - 4)

   Now, the first three terms, x² + 4x + 4, form a perfect square trinomial, which can be factored as (x + 2)².

3. Complete the square for the y terms:

   Similarly, for the expression y² - 8y, we need to add and subtract the square of half the coefficient of the y term. The coefficient of the y term is -8, so half of it is -4, and the square of -4 is 16. We add and subtract 16 within the parentheses:

   (y² - 8y + 16 - 16)

   The first three terms, y² - 8y + 16, form a perfect square trinomial, which can be factored as (y - 4)².

4. Rewrite the equation:

   Now, we substitute the perfect square trinomials back into the equation:

   (x² + 4x + 4) - 4 + (y² - 8y + 16) - 16 = -11

   This can be rewritten as:

   (x + 2)² - 4 + (y - 4)² - 16 = -11

5. Isolate the squared terms and constants:

   Next, we move the constant terms to the right side of the equation:

   (x + 2)² + (y - 4)² = -11 + 4 + 16

   Simplifying the right side, we get:

   (x + 2)² + (y - 4)² = 9

   Now, the equation is in the standard form of a circle equation.

Identifying the Center: Extracting the Coordinates

With the equation now in the standard form, (x + 2)² + (y - 4)² = 9, we can easily identify the center of the circle. Recall that the standard form is:

(x - h)² + (y - k)² = r²

where (h, k) is the center of the circle.

Comparing our equation to the standard form, we can see that:

• (x + 2)² corresponds to (x - h)², which means h = -2.
• (y - 4)² corresponds to (y - k)², which means k = 4.

Therefore, the center of the circle is (-2, 4). The right side of the equation, 9, represents r², so the radius of the circle is √9 = 3.

Conclusion: The Center Revealed and the Power of Standard Form

Through the process of completing the square, we successfully transformed the equation x² + y² + 4x - 8y + 11 = 0 into the standard form (x + 2)² + (y - 4)² = 9. This transformation allowed us to readily identify the center of the circle as (-2, 4). The ability to manipulate equations into standard forms is a powerful tool in mathematics, providing a clear and concise way to extract key information about geometric shapes and their properties.

This exercise demonstrates the importance of understanding the standard form of a circle equation and mastering the technique of completing the square. These skills are not only essential for solving geometric problems but also for building a strong foundation in mathematical reasoning and problem-solving strategies. By practicing these techniques, you can confidently tackle a wide range of mathematical challenges.

The question asks for the center of the circle whose equation is $x^2+y^2+4x-8y+11=0$. To find the center, we need to rewrite the equation in the standard form of a circle, which is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

1. Group x and y terms: $(x^2 + 4x) + (y^2 - 8y) = -11$

2. Complete the square for x terms: Take half of the coefficient of x (which is 4), square it (2^2 = 4), and add it to both sides. $(x^2 + 4x + 4) + (y^2 - 8y) = -11 + 4$

3. Complete the square for y terms: Take half of the coefficient of y (which is -8), square it ((-4)^2 = 16), and add it to both sides. $(x^2 + 4x + 4) + (y^2 - 8y + 16) = -11 + 4 + 16$

4. Rewrite as squared terms: $(x + 2)^2 + (y - 4)^2 = 9$

Now the equation is in the standard form. The center of the circle is $(-2, 4)$.

Therefore, the correct answer is A. (-2, 4).