Understanding the Circle Equation and Finding its Center and Radius

When dealing with circle equations, it's essential to know how to derive the center and radius. This tutorial will guide you through the process of converting a given circle equation into its standard form using the method of completing the square. We'll explore several examples and discuss the importance of each step in solving such problems.

The general form of a circle's equation is:

x2 y2 - 6y 0

Step 1: Completing the Square for the y-Terms

To convert this equation to its standard form, we need to complete the square for the terms involving y. Let's follow the steps:

Begin with the equation: x2 y2 - 6y 0 Move the constant terms to the right side of the equation: x2 - 6y -y2 Add 9 to both sides to complete the square: x2 - 6y 9 -y2 9 Notice that on the left side, we have x2 (y - 3)2 9

Now the equation represents a circle in its standard form, which is:

(x - a)2 (y - b)2 r2

Step 2: Identifying the Center and Radius

From the standard form, we can clearly identify the center and radius:

The center of the circle is at point (a, b), and the radius is r.

In this example:

The center is (0, 3). The radius is 3.

Examples and Practice

Let's apply the steps to another example:

x2y2 - 6y 0

Move the constant term to the right side: x2y2 - 6y 0 Add (6/2)2 9: x2(y 2 - 6y 9) 9 This simplifies to: x2(y - 3)2 9 Divide both sides by x2: (y - 3)2 9/x2

However, to keep it simple and match the standard form, we should adjust it:

Move the constant term to the right side: x2(y - 3)2 9 Divide by 9: (y - 3)2 9 So, the center is (0, 3) and the radius is 3.

Understanding and practicing these steps is crucial for solving similar problems in calculus, geometry, and algebra. The ability to manipulate and interpret circle equations is a valuable skill that can be applied to various real-world scenarios.

Conclusion

By mastering the process of completing the square and converting a given equation into its standard form, you can efficiently determine the center and radius of any circle equation. This method is not only useful for academic purposes but also in practical applications such as engineering, physics, and architecture. Happy solving!