Lesson 6.3: The Circle

Introduction

In this lesson, we will explore the fascinating world of circles, a fundamental shape in geometry. We will learn how to express circles mathematically using the Cartesian plane, focusing on two key forms of the circle's equation: center-radius form and general form. We will also dive into the concepts of completing the square to obtain the center and radius, the important relationship between a radius and a tangent at any given point on the circle, and how to transition effortlessly between these equations.

Learning Objectives

By the end of this lesson, students, you will be able to:

Write the equation of a circle in center and radius form and general form.
Use completing the square to find the center and radius of a circle.
Understand the relationship between a tangent and the radius of a circle at a point on its circumference.
Write the equation of a circle given its center and radius.
Determine the center and radius of a circle from its general equation.

The Equation of a Circle in Center-Radius Form

A circle in the Cartesian plane can be defined by its center and radius. The standard form of the equation of a circle with center at point $(h, k)$ and radius $r$ is given by:

(x - h)^2 + (y - k)^2 = r^2

Understanding the Equation

In the equation above:

$(h, k)$ represents the coordinates of the center of the circle.
$r$ is the radius of the circle, which is always a positive number.
$(x, y)$ represents any point on the circumference of the circle.

Worked Example 1

Problem: Write the equation of a circle with center at $(3, -2)$ and radius $5$.

Solution: Using the center-radius form:

Substitute $h = 3$, $k = -2$, and $r = 5$ into the standard equation.
The equation becomes:

(x - 3)^2 + (y + 2)^2 = 5^2

Simplifying this gives:

(x - 3)^2 + (y + 2)^2 = 25

Therefore, the required equation is:

(x - 3)^2 + (y + 2)^2 = 25

The General Form of the Circle

The general form of the equation of a circle can be expressed as:

Ax^2 + Ay^2 + Bx + Cy + D = 0

where $A$, $B$, $C$, and $D$ are constants, and $A$ should be positive and equal to $1$ (otherwise, you can divide through by $A$).

The general form can be derived from the center-radius form by expanding the squares:

Start with:

(x - h)^2 + (y - k)^2 = r^2

Expanding the left-hand side:

x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2

Rearranging gives:

x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0

Thus, coefficients for a general form can be identified:

$A = 1$
$B = -2h$
$C = -2k$
$D = h^2 + k^2 - r^2$

Worked Example 2

Problem: Convert the circle's equation $(x - 1)^2 + (y + 4)^2 = 9$ to general form.

Solution:

Expand both squares:

(x - 1)^2 = x^2 - 2x + 1

(y + 4)^2 = y^2 + 8y + 16

Substitute back into the equation:

x^2 - 2x + 1 + y^2 + 8y + 16 = 9

Combine like terms and move $9$ to the left-hand side:

x^2 + y^2 - 2x + 8y + 1 + 16 - 9 = 0

This simplifies to:

x^2 + y^2 - 2x + 8y + 8 = 0

Thus, the general form of the equation is:

x^2 + y^2 - 2x + 8y + 8 = 0

Completing the Square to Find the Center and Radius

Sometimes, you will be given the general form of a circle's equation and asked to determine the center and radius. This can be achieved through a process known as completing the square.

Steps to Completing the Square

Rearrange the equation into the form $Ax^2 + Ay^2 + Bx + Cy + D = 0$.
Isolate the terms containing $x$ and $y$.
Complete the square for the $x$ terms, and complete the square for the $y$ terms.
Rewrite the equation in the center-radius form.

Worked Example 3

Problem: Find the center and radius of the circle represented by the equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

Solution:

Rearranging the equation:

x^2 - 6x + y^2 + 4y = 12

For the $x$ terms, complete the square:

Take half of $-6$, square it: $(-3)^2 = 9$.
Add and subtract $9$ within the equation.

For the $y$ terms, complete the square:

Take half of $4$, square it: $(2)^2 = 4$.
Add and subtract $4$ within the equation.

The equation becomes:

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4

This simplifies to:

(x - 3)^2 + (y + 2)^2 = 25

Now we can identify the center and radius:

Center: $(3, -2)$
Radius: $5$ (since $r = \sqrt{25}$).

Relationship Between a Tangent and the Radius

At any point on the circumference of a circle, the radius drawn to that point is perpendicular to the tangent line at that point. This means that if you were to graph a circle and draw a tangent line at any point on the circle, it would meet the radius at a right angle (90 degrees).

Conceptual Understanding

This relationship is fundamental in coordinate geometry and has significant implications in calculus, especially in the study of derivatives and slopes. If $(x_1, y_1)$ is a point on the circle, then the radius at this point is calculated using:

\text{slope of the radius} = $\frac{y_1 - k}{x_1 - h}$

The slope of the tangent line at this point would then be the negative reciprocal:

\text{slope of the tangent} = -\frac{1}{\text{slope of the radius}}

This property is useful for calculating the equation of the tangent line at a given point on the circle.

Conclusion

Understanding circles is essential as they represent one of the simplest and most symmetrical shapes in geometry. Mastery of the equations of circles, the ability to convert between forms, and the comprehension of the relationship between their radius and tangents prepare you, students, for more advanced mathematical concepts. We have learned how to define a circle using various forms of its equation, and we have manipulated these equations to determine critical properties like the center and radius.

Study Notes

The center-radius form of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
The general form is $x^2 + y^2 + Bx + Cy + D = 0$.
Completing the square allows finding the center and radius from the general form.
The relationship between the radius and tangent is that the radius is perpendicular to the tangent line at the point of contact.
In problems, identify the center and radius from given equations using algebraic manipulation.