Completing the Square to Find the Center and Radius of a Circle
This article will guide you through the process of finding the center and radius of a circle given its equation. We will use completing the square to transform the equation into the standard form of a circle. This method is a powerful tool in understanding the geometry and properties of circles, which is particularly useful in fields like engineering, physics, and mathematics.
Understanding the Standard Form of a Circle
The standard form of a circle's equation is ( (x - h)^2 (y - k)^2 r^2 ). Here, ((h, k)) represents the center of the circle, and (r) represents the radius. Our goal is to convert the given equation into this standard form by completing the square.
Example 1: Find the Center and Radius of (x^2 y^2 - 6y - 27 0)
Given the equation: (x^2 y^2 - 6y - 27 0)
Rearrange the equation:
(x^2 y^2 - 6y 27)
Complete the square for the (y) terms:
- The coefficient of (y) is (-6). Halve it and square it to get (-3) and (9):
(x^2 y^2 - 6y 9 - 9 27)
Combine and simplify:
(x^2 (y - 3)^2 - 9 27)
(x^2 (y - 3)^2 36)
Identify the center and radius:
The equation is now in the standard form ((x - h)^2 (y - k)^2 r^2), where (h 0) and (k 3), and (r sqrt{36} 6).
Therefore, the center is ((0, 3)) and the radius is (6).
Example 2: Find the Center and Radius of (x^2 y^2 - 6x - 12y - 55 0)
Given the equation: (x^2 y^2 - 6x - 12y - 55 0)
Rearrange the equation:
(x^2 - 6x y^2 - 12y 55)
Complete the square for the (x) terms and (y) terms:
- For (x^2 - 6x), add and subtract (9):
(x^2 - 6x 9 - 9)
- For (y^2 - 12y), add and subtract (36):
(y^2 - 12y 36 - 36)
Combine and simplify:
((x^2 - 6x 9) (y^2 - 12y 36) - 9 - 36 55)
((x - 3)^2 (y - 6)^2 - 45 55)
((x - 3)^2 (y - 6)^2 100)
Identify the center and radius:
This equation is now in the standard form ((x - h)^2 (y - k)^2 r^2), where (h 3), (k 6), and (r sqrt{100} 10).
Therefore, the center is ((3, 6)) and the radius is (10).
Example 3: Find the Center and Radius of (x^2 y^2 - 6x - 12y 55)
Given the equation: (x^2 y^2 - 6x - 12y 55)
Rearrange the equation:
(x^2 - 6x y^2 - 12y 55)
Complete the square for the (x) terms and (y) terms:
- For (x^2 - 6x), add and subtract (9):
(x^2 - 6x 9 - 9)
- For (y^2 - 12y), add and subtract (36):
(y^2 - 12y 36 - 36)
Combine and simplify:
((x^2 - 6x 9) (y^2 - 12y 36) - 9 - 36 55)
((x - 3)^2 (y - 6)^2 - 45 55)
((x - 3)^2 (y - 6)^2 100)
Identify the center and radius:
This equation is now in the standard form ((x - h)^2 (y - k)^2 r^2), where (h 3), (k 6), and (r sqrt{100} 10).
Therefore, the center is ((3, 6)) and the radius is (10).
Conclusion
Completing the square is a powerful algebraic technique that helps us transform the equation of a circle into the standard form. This transformation is crucial for understanding the geometric properties of the circle, such as its center and radius. The steps involve rearranging the equation, completing the square for each variable, and then identifying the standard form.
Understanding this process not only simplifies solving problems but also enhances your analytical skills in mathematics. Whether you are a student preparing for exams or a professional dealing with practical applications, the ability to work with circle equations is invaluable.