Equation of a Circle Tangential to Both Axes at Equal Distance from Origin

Introduction

Understanding the equation of a circle that is tangential to both axes at a specified distance from the origin is a fascinating topic. This article explores the concept in depth, providing a clear understanding of the mathematical principles involved. We will also delve into the different cases where the circle can be located in various quadrants.

Conceptual Overview

A circle is said to be tangential to both axes if it touches the x-axis and the y-axis at a certain distance from the origin. In this particular instance, the distance is 4 units. This means the circle will have its tangency points at (4,0), (0,4), (-4,0), and (0,-4).

General Formula for a Circle

The equation of a circle with center at (a, b) and radius r is given by:

[ (x - a)^2 (y - b)^2 r^2 ]

Case 1: First Quadrant

Center at (4, 4)

In the first quadrant, the circle is centered at (4, 4) and has a radius of 4 units. The equation of the circle is:

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

This can be expanded to:

[ x^2 - 8x 16 y^2 - 8y 16 16 ]

Simplifying further:

[ x^2 y^2 - 8x - 8y 16 0 ]

Case 2: Second Quadrant

Center at (-4, 4)

In the second quadrant, the circle is centered at (-4, 4). The equation of the circle is:

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

This can be expanded to:

[ x^2 8x 16 y^2 - 8y 16 16 ]

Simplifying further:

[ x^2 y^2 8x - 8y 16 0 ]

Case 3: Third Quadrant

Center at (-4, -4)

In the third quadrant, the circle is centered at (-4, -4). The equation of the circle is:

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

This can be expanded to:

[ x^2 8x 16 y^2 8y 16 16 ]

Simplifying further:

[ x^2 y^2 8x 8y 16 0 ]

Case 4: Fourth Quadrant

Center at (4, -4)

In the fourth quadrant, the circle is centered at (4, -4). The equation of the circle is:

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

This can be expanded to:

[ x^2 - 8x 16 y^2 8y 16 16 ]

Simplifying further:

[ x^2 y^2 - 8x 8y 16 0 ]

The General Case

In a more general sense, if the circle is tangential to both the x-axis and the y-axis, the coordinates of the center of the circle will be equal in magnitude and opposite in sign. Therefore, the radius r is given by:

[ r sqrt{x^2 y^2} sqrt{2x} ]

Thus, the equation of the circle is:

[ (x - a)^2 (y - b)^2 2a ]

Since a b:

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

Conclusion

By understanding the equations and the placement of the circle in different quadrants, we can deduce the specific equations for each case. This knowledge is valuable in various applications of geometry and calculus.