Circle Equation: Radius 2, Shared Center

Finding the equation of a circle given its radius and center coordinates is a fundamental concept in coordinate geometry. In this problem, we're tasked with determining the equation of a circle that has a radius of 2 units and shares its center with another circle whose equation is given as $x^2+y^2-8x-6y+24=0$. This involves completing the square to find the center of the given circle and then using the standard form of a circle's equation to find the required equation. Let's dive into a detailed explanation of how to solve this problem.

Understanding the Problem

Before we begin, let's recap the key information:

• Radius of the target circle: 2 units
• Center of the target circle: Same as the center of the circle defined by $x^2+y^2-8x-6y+24=0$

Our goal is to find the equation of the circle in the standard form: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle, and $r$ is the radius.

Step 1: Finding the Center of the Given Circle

The equation of the given circle is $x^2+y^2-8x-6y+24=0$. To find its center, we need to rewrite this equation in the standard form by completing the square for both $x$ and $y$ terms.

Completing the Square for x

We have $x^2 - 8x$. To complete the square, we need to add and subtract $(\frac{-8}{2})^2 = (-4)^2 = 16$. So, $x^2 - 8x = (x^2 - 8x + 16) - 16 = (x - 4)^2 - 16$.

Completing the Square for y

We have $y^2 - 6y$. To complete the square, we need to add and subtract $(\frac{-6}{2})^2 = (-3)^2 = 9$. So, $y^2 - 6y = (y^2 - 6y + 9) - 9 = (y - 3)^2 - 9$.

Rewriting the Circle Equation

Now, substitute these back into the original equation: $(x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0$ $(x - 4)^2 + (y - 3)^2 - 16 - 9 + 24 = 0$ $(x - 4)^2 + (y - 3)^2 - 1 = 0$ $(x - 4)^2 + (y - 3)^2 = 1$

From this standard form, we can see that the center of the given circle is $(4, 3)$.

Step 2: Forming the Equation of the Target Circle

Now that we know the center of the target circle is $(4, 3)$ and the radius is 2, we can write the equation of the target circle using the standard form $(x - h)^2 + (y - k)^2 = r^2$.

Here, $(h, k) = (4, 3)$ and $r = 2$. So, the equation is $(x - 4)^2 + (y - 3)^2 = 2^2$, which simplifies to $(x - 4)^2 + (y - 3)^2 = 4$.

Conclusion

The equation of the circle with a radius of 2 units and the same center as the circle defined by $x^2+y^2-8x-6y+24=0$ is $(x - 4)^2 + (y - 3)^2 = 4$. Comparing this with the given options, we find that option C, $(x-4)^2+(y-3)^2=2^2$, is the correct answer. This problem showcases the importance of understanding the standard form of a circle's equation and the technique of completing the square. By mastering these concepts, you can solve a wide range of circle-related problems in coordinate geometry. Remember, completing the square allows us to transform general quadratic equations into a form that reveals key information, such as the center and radius of a circle. This skill is not only valuable for solving academic problems but also for various applications in engineering, physics, and computer graphics, where understanding geometric shapes and their equations is crucial. Keep practicing, and you'll become more proficient in solving these types of problems! Good luck!

Detailed Explanation of Circle Equations

To deeply understand how we arrived at the solution, let's discuss the fundamental concepts of circle equations. A circle in a two-dimensional Cartesian plane is defined as the set of all points equidistant from a central point. The standard form equation of a circle is expressed as:

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

Where:

• $(x, y)$ represents any point on the circle's circumference.
• $(h, k)$ represents the coordinates of the center of the circle.
• $r$ represents the radius of the circle, which is the distance from the center to any point on the circumference.

Understanding the Components

• Center (h, k): The center is the pivotal point around which the circle is drawn. Changing the values of h and k shifts the circle's position on the Cartesian plane. For example, a circle centered at (0,0) is at the origin, while a circle centered at (3, -2) is shifted 3 units to the right and 2 units down from the origin.
• Radius (r): The radius determines the size of the circle. It is the distance from the center to any point on the circle. The larger the radius, the larger the circle. The radius must always be a positive value because it represents a distance.
• $(x - h)^2 + (y - k)^2$: This part of the equation represents the square of the distance between any point (x, y) on the circle and the center (h, k). According to the Pythagorean theorem, this distance must equal the radius r.

General Form of a Circle Equation

While the standard form is highly useful for identifying the center and radius, a circle's equation can also be expressed in the general form:

$x^2 + y^2 + Ax + By + C = 0$

Where A, B, and C are constants. This form is less intuitive for directly reading off the center and radius, but it is often encountered in various mathematical contexts. To convert from the general form to the standard form, you typically need to complete the square for both the x and y terms, as demonstrated in the problem above. Completing the square involves rearranging the equation and adding specific constants to create perfect square trinomials, which can then be factored into the form $(x - h)^2$ and $(y - k)^2$.

Completing the Square: A Detailed Look

Completing the square is a technique used to rewrite a quadratic expression in a more convenient form. For a quadratic expression of the form $x^2 + bx$, we add and subtract $(\frac{b}{2})^2$ to complete the square:

$x^2 + bx = x^2 + bx + (\frac{b}{2})^2 - (\frac{b}{2})^2 = (x + \frac{b}{2})^2 - (\frac{b}{2})^2$

This allows us to rewrite the expression as a squared term plus a constant. In the context of circle equations, we apply this technique to both the x and y terms to transform the general form into the standard form.

Example

Let's consider the general form equation:

$x^2 + y^2 - 4x + 6y - 3 = 0$

To convert this to standard form:

1. Group x and y terms: $(x^2 - 4x) + (y^2 + 6y) = 3$
2. Complete the square for x: $x^2 - 4x + (\frac{-4}{2})^2 = x^2 - 4x + 4 = (x - 2)^2$
3. Complete the square for y: $y^2 + 6y + (\frac{6}{2})^2 = y^2 + 6y + 9 = (y + 3)^2$
4. Add the completed square constants to both sides of the equation: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 3 + 4 + 9$
5. Rewrite as squared terms: $(x - 2)^2 + (y + 3)^2 = 16$

Now, the equation is in standard form, and we can easily identify the center as (2, -3) and the radius as $\sqrt{16} = 4$.

Applications and Importance

Understanding circle equations is crucial not only for academic problem-solving but also for various real-world applications. Some examples include:

• Computer Graphics: In computer graphics, circles and circular arcs are fundamental building blocks for creating complex shapes and animations. Understanding their equations allows developers to accurately render and manipulate these shapes on the screen.
• Engineering: In mechanical engineering, circles are used in the design of gears, wheels, and other circular components. Understanding their properties is essential for ensuring proper functionality and performance.
• Physics: In physics, circular motion is a fundamental concept. Understanding circle equations helps in analyzing the motion of objects moving in circular paths, such as satellites orbiting the Earth.
• Navigation: Circle equations are used in navigation systems to determine the position of a vehicle or aircraft based on the signals received from multiple sources.

Tips for Solving Circle Equation Problems

1. Memorize the standard and general forms of the circle equation.
2. Practice completing the square. This technique is essential for converting between the general and standard forms.
3. Always check your work. Make sure that the center and radius you identify from the standard form match the given information in the problem.
4. Visualize the circle. Sketching a rough diagram can help you understand the problem and identify potential errors.
5. Pay attention to the signs. A common mistake is to confuse the signs of the center coordinates in the standard form.

By mastering these concepts and practicing regularly, you'll become more confident in solving circle equation problems and applying them to various real-world scenarios. Remember, mathematics is a building block, so each concept you learn builds upon the previous ones. Keep exploring, keep practicing, and you'll achieve success!