Introduction to Circle Equation: Finding the Equation of a Circle Tangent to a Given Line

Understanding the intricacies of geometric shapes and their relationships can be fascinating, especially when it comes to circles. In this article, we will explore how to find the equation of a circle with a specified center that is tangent to a given line. We will walk through a detailed step-by-step process, employing geometric and algebraic tools to derive the desired equation. This content is essential for students and professionals alike interested in geometry, calculus, and more.

Step-by-Step Process

Let’s start with the problem: Find the equation of the circle with the center at ((-4, 2)) and tangent to the line (2x - y 2 0).

Identify the Center of the Circle

The center of the circle is given as ((-4, 2)).

Calculate the Distance from the Center to the Line

The formula to find the distance (d) from a point ((x_0, y_0)) to a line (Ax By C 0) is:

[d frac{|Ax_0 By_0 C|}{sqrt{A^2 B^2}}]

For the line (2x - y 2 0), the coefficients are:

(A 2) (B -1) (C 2)

The center of the circle is ((-4, 2)), so substituting these values, we get:

[d frac{|2(-4) - 1(2) 2|}{sqrt{2^2 (-1)^2}} frac{|-8 - 2 2|}{sqrt{4 1}} frac{|-8|}{sqrt{5}} frac{8}{sqrt{5}} frac{8sqrt{5}}{5}]

Determine the Radius of the Circle

Since the circle is tangent to the line, the radius (r) of the circle is equal to the distance (d):

[r frac{8sqrt{5}}{5}]

Write the Equation of the Circle

The standard form of the equation of a circle with center ((h, k)) and radius (r) is:

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

Substituting the center ((-4, 2)) and the radius (r frac{8sqrt{5}}{5}), we get:

[ (x 4)^2 (y - 2)^2 left(frac{8sqrt{5}}{5}right)^2 ]

Calculating (r^2):

[ r^2 left(frac{8sqrt{5}}{5}right)^2 frac{64 cdot 5}{25} frac{320}{25} 12.8 ]

Therefore, the equation of the circle is:

[ (x 4)^2 (y - 2)^2 12.8 ]

Additional Insights

In the course of our derivation, we have utilized the distance formula to find the radius, and then have used the standard form of the circle equation to write the final answer. This problem intertwines concepts from both geometry (distance and circle properties) and algebra (solving equations).

Let's break down further by considering the algebraic approach:

The distance from the center ((-4, 2)) to the line (2x - y 2 0) can also be stated as the radius (R) of the circle:

[ text{Distance} R frac{-4 - 2 2}{sqrt{5}} frac{-4}{sqrt{5}} frac{4sqrt{5}}{5} ]

The equation for the radius being the distance from the center to the line can be used to derive the equation of the circle:

[ (x 4)^2 (y - 2)^2 left(frac{4sqrt{5}}{5}right)^2 frac{16}{5} ]

This confirms that the equation of the circle is:

[ (x 4)^2 (y - 2)^2 frac{16}{5} ]

This expression can be rephrased to show the relationship between the center and the line, ultimately arriving at the same equation we obtained earlier.

Visualizing the Circle and the Line

A visual representation of the circle and the tangent line can be helpful for understanding the geometric configuration. Using graphing tools or software, we can plot the circle with center ((-4, 2)) and radius (frac{8sqrt{5}}{5}), and the line (2x - y 2 0). The point where the circle touches the line is a point of tangency, and this intersection confirms the equation we derived.

Conclusion

By following the steps detailed in this article, we have successfully derived the equation of a circle with the given center and tangent to the specified line. The key steps involved identifying the center, calculating the distance (which is the radius), and then using the circle’s standard form to write the equation. This process not only solves the problem but also reinforces the principles of geometry and algebra.

Feel free to explore and apply these concepts in various scenarios and problems involving circles and lines. For further reading and exercises, consider using textbooks, online resources, and practice problems to solidify your understanding.