Introduction to Circle Equations and X-Axis Contact

In this article, we will explore how to find the equation of a circle that touches the x-axis and has a specific center. This type of problem is common in geometry and can be solved using basic algebraic methods. Let's dive into the details.

Key Concepts and Notations

For a circle, there are a few fundamental concepts to understand:

Center: The center of a circle, often denoted as (h, k). Radius: The distance from the center to any point on the circumference of the circle.

The general formula for the equation of a circle is:

General Equation of a Circle

Equation: [(x - h)^2 (y - k)^2 r^2]

Here, ((h, k)) represents the center of the circle, and (r) is the radius.

Solving the Given Problem

Consider a circle that touches the x-axis at the point 23. Since it touches the x-axis, its center is 3 units above the x-axis. Therefore, the radius of the circle is 3.

Step 1: Identify the Center and Radius

The center of the circle is at the point (2, 3). The radius of the circle is 3 units.

Step 2: Apply the General Equation of a Circle

Substituting the center ((h, k) (2, 3)) and the radius (r 3) into the general equation of a circle, we get:

[(x - 2)^2 (y - 3)^2 3^2]

This simplifies to:

[(x - 2)^2 (y - 3)^2 9]

Step 3: Simplify and Verify

Expanding the equation:

[(x - 2)^2 x^2 - 4x 4]

[(y - 3)^2 y^2 - 6y 9]

Combining these terms:

[(x^2 - 4x 4) (y^2 - 6y 9) 9]

This simplifies to:

[x^2 - 4x y^2 - 6y 9 9]

Subtracting 9 from both sides:

[x^2 - 4x y^2 - 6y 0]

Conclusion

The equation of the circle that touches the x-axis at 23 and has a center at (2, 3) is:

[x^2 - 4x y^2 - 6y 0]

This can be written as:

[(x - 2)^2 (y - 3)^2 9]

Using the general formula, this confirms our steps and the solution is verified.