Equation Of A Circle Radius 2 And Common Center Detailed Solution
In mathematics, a circle is defined as the set of all points in a plane that are equidistant from a central point. The equation of a circle is a fundamental concept in coordinate geometry, allowing us to describe circles algebraically. This article delves into determining the equation of a circle given its radius and center, where the center is defined by another circle's equation. We will explore the standard form of a circle's equation, methods for finding the center of a circle from its general equation, and then apply these concepts to solve the problem at hand. The goal is to provide a comprehensive understanding of how to find the equation of a circle based on given parameters, making it easy for anyone to grasp the underlying principles and apply them to similar problems. Understanding the relationship between the equation and the geometric properties of a circle is crucial in various fields, including geometry, physics, and engineering. This article aims to bridge the gap between theoretical knowledge and practical application, offering clear explanations and step-by-step solutions to enhance your problem-solving skills.
Understanding the Standard Equation of a Circle
The standard equation of a circle is a powerful tool for representing circles in the Cartesian plane. It provides a clear and concise way to describe a circle's properties, such as its center and radius. The standard form is given by:
(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 equation arises from the Pythagorean theorem, applied to any point (x, y) on the circle, the center (h, k), and the radius r. The distance between any point on the circle and the center is always equal to the radius. By understanding this equation, we can easily determine the center and radius of a circle if its equation is given in standard form, and conversely, we can write the equation of a circle if we know its center and radius.
The beauty of the standard form lies in its simplicity and directness. It allows us to immediately visualize the circle on a coordinate plane. For example, if we have the equation (x – 2)² + (y + 3)² = 9, we can quickly identify that the center of the circle is (2, -3) and the radius is √9 = 3. This makes the standard form invaluable for solving problems related to circles, such as finding the equation of a circle given certain conditions or determining the intersection points of circles and lines. Furthermore, the standard form serves as a foundation for more advanced concepts in analytic geometry, such as conic sections and their properties. Mastering the standard equation of a circle is therefore essential for anyone studying mathematics or related fields.
Finding the Center of a Circle from its General Equation
While the standard form of a circle's equation is highly informative, circles are often represented in the general form:
x² + y² + Dx + Ey + F = 0
where D, E, and F are constants. To find the center of a circle given its equation in general form, we need to convert it into the standard form. This is achieved by completing the square for both the x and y terms. The process involves rearranging the equation, grouping the x terms and y terms together, and then adding and subtracting appropriate constants to create perfect square trinomials. This technique allows us to rewrite the general equation in the standard form (x – h)² + (y – k)² = r², from which the center (h, k) can be easily identified.
Completing the square is a fundamental algebraic technique that transforms a quadratic expression into a perfect square trinomial plus a constant. This technique is crucial not only for finding the center of a circle but also for solving quadratic equations and simplifying various algebraic expressions. To complete the square for an expression of the form x² + bx, we add and subtract (b/2)² to the expression. This creates a perfect square trinomial (x + b/2)² minus the constant (b/2)². Applying this method to both the x and y terms in the general equation allows us to rewrite it in the standard form, revealing the center and radius of the circle. The ability to convert between the general and standard forms is essential for solving a wide range of problems involving circles and other conic sections.
For example, consider the equation x² + y² – 4x + 6y – 12 = 0. To find the center, we first group the x and y terms: (x² – 4x) + (y² + 6y) = 12. Then, we complete the square for the x terms by adding and subtracting (-4/2)² = 4, and for the y terms by adding and subtracting (6/2)² = 9. This gives us (x² – 4x + 4) + (y² + 6y + 9) = 12 + 4 + 9, which simplifies to (x – 2)² + (y + 3)² = 25. From this standard form, we can see that the center of the circle is (2, -3). This process highlights the importance of mastering algebraic techniques like completing the square for solving geometric problems.
Problem Statement and Solution
Now, let's address the specific problem: Find the equation of a circle with a radius of 2 units, where the center is the same as the center of the circle whose equation is:
x² + y² – 8x – 6y + 24 = 0
Step 1: Find the Center of the Given Circle
First, we need to find the center of the circle described by the equation x² + y² – 8x – 6y + 24 = 0. To do this, we will convert the general form of the equation to the standard form by completing the square. This process involves rearranging the terms, grouping the x and y terms, and then adding and subtracting the appropriate constants to create perfect square trinomials.
Group the x and y terms:
(x² – 8x) + (y² – 6y) = -24
Complete the square for the x terms. To do this, take half of the coefficient of the x term (-8), square it ((-4)² = 16), and add it to both sides of the equation:
(x² – 8x + 16) + (y² – 6y) = -24 + 16
Similarly, complete the square for the y terms. Take half of the coefficient of the y term (-6), square it ((-3)² = 9), and add it to both sides of the equation:
(x² – 8x + 16) + (y² – 6y + 9) = -24 + 16 + 9
Now, rewrite the expressions in parentheses as perfect squares:
(x – 4)² + (y – 3)² = 1
From this standard form, we can identify the center of the circle as (4, 3).
Step 2: Use the Center and Radius to Form the New Equation
We are given that the circle we want to find has a radius of 2 units and the same center as the circle we just analyzed, which is (4, 3). Now we can use the standard form of the circle equation to find the equation of the desired circle. The standard form is:
(x – h)² + (y – k)² = r²
where (h, k) is the center and r is the radius. Plug in the values we have:
- h = 4
- k = 3
- r = 2
The equation becomes:
(x – 4)² + (y – 3)² = 2²
(x – 4)² + (y – 3)² = 4
Thus, the equation of the circle with a radius of 2 and the center at (4, 3) is (x – 4)² + (y – 3)² = 4.
Analyzing the Options
Now, let's compare our solution with the given options:
- (x + 4)² + (y + 3)² = 2
- (x – 4)² + (y – 3)² = 2
- (x – 4)² + (y – 3)² = 4
Our solution, (x – 4)² + (y – 3)² = 4, matches option 3. Therefore, the correct equation representing the circle described is option 3.
This problem illustrates the importance of understanding the standard equation of a circle and the technique of completing the square. By mastering these concepts, we can solve a variety of problems related to circles and their properties. The ability to convert between the general and standard forms of a circle's equation is a valuable skill in coordinate geometry and is essential for tackling more complex problems.
In conclusion, determining the equation of a circle requires a solid understanding of its properties, particularly the center and radius, and their relationship to the standard form of the circle's equation. This article has provided a detailed explanation of how to find the center of a circle from its general equation by completing the square and how to use the center and radius to write the equation in standard form. By working through the example problem, we demonstrated a step-by-step approach to solving such problems, highlighting the importance of algebraic manipulation and a clear understanding of geometric concepts. The ability to solve these types of problems is crucial in various fields, including mathematics, physics, and engineering. Mastering these techniques enhances your problem-solving skills and provides a solid foundation for more advanced topics in geometry and related disciplines. Whether you are a student learning about circles for the first time or a professional looking to refresh your knowledge, this article serves as a valuable resource for understanding and applying the concepts related to the equation of a circle.
