Finding The Center And Radius Of A Circle $x^2+y^2-6x+10y+25=0$

by THE IDEN 64 views

Determining the center and radius of a circle from its equation is a fundamental concept in coordinate geometry. This article provides a detailed, step-by-step guide on how to find the center and radius of a circle when given its equation in the general form. We will specifically address the equation x2+y2βˆ’6x+10y+25=0x^2 + y^2 - 6x + 10y + 25 = 0, offering a clear explanation suitable for students and anyone interested in refreshing their understanding of circle equations. We'll explore the underlying principles, the algebraic manipulations required, and provide a comprehensive analysis to ensure clarity and understanding. By the end of this guide, you will be equipped with the knowledge to confidently tackle similar problems and gain a deeper appreciation for the elegance of geometric equations.

Understanding the General and Standard Forms of a Circle Equation

To effectively find the center and radius, it's crucial to understand the two primary forms of a circle's equation: the general form and the standard form. The general form of a circle's equation is expressed as:

Ax2+By2+Cx+Dy+E=0Ax^2 + By^2 + Cx + Dy + E = 0

where A, B, C, D, and E are constants. In our specific case, the equation x2+y2βˆ’6x+10y+25=0x^2 + y^2 - 6x + 10y + 25 = 0 is already in this general form. However, extracting the center and radius directly from this form can be challenging. That's where the standard form comes in. The standard form, also known as the center-radius form, is given by:

(xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2

Here, (h,k)(h, k) represents the coordinates of the center of the circle, and rr represents the radius. The standard form provides a direct and intuitive way to identify the circle's center and radius. The key to solving our problem lies in transforming the given general form equation into the standard form. This transformation involves a powerful algebraic technique called completing the square. Completing the square allows us to rewrite quadratic expressions in a form that reveals the center and radius clearly. We'll delve into the mechanics of completing the square in the next section, demonstrating how it unlocks the hidden geometry within the equation.

The Power of Completing the Square: Transforming the Equation

Completing the square is the core technique we'll employ to convert the given equation x2+y2βˆ’6x+10y+25=0x^2 + y^2 - 6x + 10y + 25 = 0 into standard form. This method allows us to rewrite quadratic expressions as perfect square trinomials, which are easily factored into squared terms. Let's break down the process step-by-step. First, we group the xx terms and the yy terms together, and move the constant term to the right side of the equation:

(x2βˆ’6x)+(y2+10y)=βˆ’25(x^2 - 6x) + (y^2 + 10y) = -25

Now, we focus on completing the square for the xx terms. To do this, we take half of the coefficient of the xx term (-6), square it ((-3)^2 = 9), and add it to both sides of the equation. This ensures that we maintain the equality. We repeat this process for the yy terms, taking half of the coefficient of the yy term (10), squaring it ((5)^2 = 25), and adding it to both sides:

(x2βˆ’6x+9)+(y2+10y+25)=βˆ’25+9+25(x^2 - 6x + 9) + (y^2 + 10y + 25) = -25 + 9 + 25

Notice that the expressions inside the parentheses are now perfect square trinomials. We can factor them as follows:

(xβˆ’3)2+(y+5)2=9(x - 3)^2 + (y + 5)^2 = 9

This is now in the standard form of a circle's equation! By carefully completing the square, we've transformed the original equation into a form that readily reveals the center and radius of the circle. In the next section, we'll extract these key parameters and discuss their geometric significance.

Identifying the Center and Radius: A Geometric Interpretation

Now that we have the equation in the standard form, (xβˆ’3)2+(y+5)2=9(x - 3)^2 + (y + 5)^2 = 9, identifying the center and radius is straightforward. By comparing this equation to the general standard form (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, we can directly extract the values of hh, kk, and rr. Remember that (h,k)(h, k) represents the center of the circle, and rr represents the radius. In our equation, we have:

  • h=3h = 3
  • k=βˆ’5k = -5 (note that the equation has (y+5)(y + 5), which corresponds to (yβˆ’(βˆ’5))(y - (-5)))
  • r2=9r^2 = 9, so r=9=3r = \sqrt{9} = 3

Therefore, the center of the circle is (3,βˆ’5)(3, -5), and the radius is 3. This means that if we were to plot this circle on a coordinate plane, the central point would be located at the coordinates (3, -5), and every point on the circle would be exactly 3 units away from this center. The radius is a fundamental property of a circle, defining its size and extent. Understanding the center and radius allows us to visualize the circle's position and dimensions in the coordinate plane. In the next section, we'll summarize our findings and discuss the implications of this solution.

Solution and Conclusion: Summarizing the Circle's Properties

In summary, we started with the equation of a circle in general form, x2+y2βˆ’6x+10y+25=0x^2 + y^2 - 6x + 10y + 25 = 0. Through the process of completing the square, we transformed the equation into the standard form, (xβˆ’3)2+(y+5)2=9(x - 3)^2 + (y + 5)^2 = 9. By comparing this to the standard form equation of a circle, we were able to identify the center and radius:

  • Center: (3,βˆ’5)(3, -5)
  • Radius: 33

Therefore, the correct answer is D. Center (3,-5); radius 3. This exercise demonstrates the power of algebraic manipulation in revealing geometric properties. Completing the square is a versatile technique that extends beyond circle equations and is applicable in various areas of mathematics. Understanding the relationship between the general and standard forms of a circle's equation provides a solid foundation for further exploration of conic sections and analytic geometry. This comprehensive guide has not only provided a solution to the specific problem but also illuminated the underlying concepts and techniques involved in analyzing circle equations. By mastering these principles, you can confidently approach similar problems and deepen your understanding of the beautiful interplay between algebra and geometry.