Find The Equation Of A Circle With A Radius Of 2 And The Same Center As The Circle Defined By $x^2 + Y^2 - 8x - 6y + 24 = 0$.

by ADMIN 126 views

This article delves into the process of identifying the equation that represents a circle, given specific characteristics such as its radius and center. We will focus on a particular problem: determining the equation of a circle with a radius of 2 units, whose center coincides with the center of another circle defined by the equation x2+y28x6y+24=0x^2 + y^2 - 8x - 6y + 24 = 0. This exploration will provide a detailed walkthrough of the steps involved, ensuring a clear understanding of the underlying concepts and techniques.

Understanding the Circle Equation

Before diving into the problem, let's first solidify our understanding of the circle equation. The standard form equation of a circle with center (h, k) and radius r is:

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

This equation is derived from the Pythagorean theorem, where the distance between any point (x, y) on the circle and the center (h, k) is equal to the radius r. By understanding this fundamental equation, we can effectively manipulate and interpret circle-related problems. The key elements to identify are the center coordinates (h, k) and the radius (r), as they directly determine the circle's position and size in the coordinate plane. These concepts form the bedrock for solving more complex problems involving circles.

Determining the Center of the Given Circle

In this particular problem, we are given the equation of a circle in general form: x2+y28x6y+24=0x^2 + y^2 - 8x - 6y + 24 = 0. To find the center of this circle, we need to convert it to the standard form mentioned above. This is achieved by completing the square for both the x and y terms. Completing the square is a technique used to rewrite a quadratic expression in a form that includes a perfect square trinomial, which can then be factored into a squared binomial. The process involves taking half of the coefficient of the x term (which is -8), squaring it ((-4)^2 = 16), and adding it to both sides of the equation. We do the same for the y term: take half of the coefficient of the y term (which is -6), squaring it ((-3)^2 = 9), and adding it to both sides. By completing the square, we can rewrite the original equation in a more manageable form that reveals the center and radius of the circle. This technique is crucial in various mathematical contexts, including conic sections and calculus.

Let's rewrite the equation by grouping the x terms and y terms together:

(x28x)+(y26y)+24=0(x^2 - 8x) + (y^2 - 6y) + 24 = 0

Now, complete the square for the x terms. Take half of the coefficient of x (-8), which is -4, and square it to get 16. Add 16 to both sides of the equation:

(x28x+16)+(y26y)+24=16(x^2 - 8x + 16) + (y^2 - 6y) + 24 = 16

Next, complete the square for the y terms. Take half of the coefficient of y (-6), which is -3, and square it to get 9. Add 9 to both sides of the equation:

(x28x+16)+(y26y+9)+24=16+9(x^2 - 8x + 16) + (y^2 - 6y + 9) + 24 = 16 + 9

Now, rewrite the expressions in parentheses as squared binomials:

(x4)2+(y3)2+24=25(x - 4)^2 + (y - 3)^2 + 24 = 25

Subtract 24 from both sides to isolate the squared terms:

(x4)2+(y3)2=1(x - 4)^2 + (y - 3)^2 = 1

From this standard form, we can clearly see that the center of the circle is (4, 3). The right side of the equation, 1, represents the square of the radius of this circle. This means the radius of the given circle is √1 = 1 unit. However, the problem specifies that the circle we are looking for has a radius of 2 units and the same center. Understanding how to extract the center and radius from both the general and standard forms of a circle equation is essential for solving problems related to circles and their properties. This process involves algebraic manipulation, including completing the square, which is a powerful technique in mathematics.

Constructing the Equation of the Described Circle

Now that we know the center of the circle is (4, 3) and the radius is 2 units, we can construct the equation of the circle. We'll use the standard form equation: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. Remember, (h, k) represents the center, and r represents the radius. Substituting the values we have, we get:

(x4)2+(y3)2=22(x - 4)^2 + (y - 3)^2 = 2^2

Simplifying the equation, we have:

(x4)2+(y3)2=4(x - 4)^2 + (y - 3)^2 = 4

This is the equation of the circle with a center at (4, 3) and a radius of 2 units. We have successfully constructed the equation based on the given parameters, demonstrating the direct application of the standard form equation of a circle. The process of constructing an equation from given characteristics is fundamental in geometry and allows us to mathematically represent geometric figures accurately. This skill is crucial for various applications in fields like engineering, physics, and computer graphics.

Analyzing the Answer Choices

Now, let's analyze the provided answer choices to identify the correct equation:

A. (x+4)2+(y+3)2=2(x + 4)^2 + (y + 3)^2 = 2 B. $(x - 4)^2 + (y - 3)^2 = 4

Comparing these options with the equation we derived, (x4)2+(y3)2=4(x - 4)^2 + (y - 3)^2 = 4, it is clear that option B matches our result. Option A has incorrect signs within the parentheses and an incorrect radius squared value, making it an invalid choice. The ability to analyze and compare equations is a critical skill in mathematics. It allows us to verify solutions, identify errors, and ultimately, gain a deeper understanding of the mathematical relationships involved. In this context, careful comparison of the derived equation with the given options ensures that we select the correct representation of the described circle.

Conclusion: The Correct Equation

Therefore, the equation that represents the circle with a radius of 2 units and a center at (4, 3) is:

B. (x4)2+(y3)2=4(x - 4)^2 + (y - 3)^2 = 4

This problem highlights the importance of understanding the standard form equation of a circle and the technique of completing the square. By systematically applying these concepts, we can effectively solve problems involving circles and their equations. Mastering these fundamental skills is crucial for success in mathematics and related fields. The process of deriving and verifying equations not only strengthens our understanding of geometric shapes but also enhances our problem-solving abilities in a broader mathematical context.

Keywords for SEO Optimization:

To ensure this article is easily discoverable by users searching for related information, we have incorporated the following keywords:

  • Circle equation: This is the most fundamental keyword, as it directly relates to the topic at hand.
  • Standard form of circle equation: This specifies the format of the equation we are working with.
  • Center of a circle: Identifying the center is a key step in solving the problem.
  • Radius of a circle: The radius is another crucial parameter in defining a circle.
  • Completing the square: This is the algebraic technique used to find the center and radius.
  • Equation of a circle with given center and radius: This phrase accurately describes the problem we are solving.
  • Mathematics: This broad keyword categorizes the topic.
  • Geometry: This is a more specific area of mathematics that deals with shapes and their properties.
  • Conic sections: Circles are a type of conic section.
  • Algebra: The algebraic manipulations involved in solving the problem are relevant.

By strategically including these keywords throughout the article, we increase its visibility in search engine results, making it a valuable resource for individuals seeking information about circle equations and related topics. The use of both broad and specific keywords ensures that the article can be found by a wide range of users, from students learning the basics of circle equations to individuals seeking more advanced mathematical concepts. This approach maximizes the article's reach and impact.