Section 2.2: Circles – Equations, Properties, and Graphing (College Algebra Study Notes)

Circles in College Algebra

Solving Quadratic Equations by Completing the Square

Completing the square is a fundamental algebraic technique used to solve quadratic equations and is essential for deriving the equation of a circle. The process involves manipulating a quadratic equation into a perfect square trinomial.

• Definition: Completing the square means rewriting a quadratic expression of the form $x^2 + bx$ as $(x + d)^2 + c$.

• Key Step: To complete the square for $x^2 + 10x$, add $(\frac{10}{2})^2 = 25$.

• Example: For $x^2 + 10x$, half of 10 is 5, and $5^2 = 25$.

• Application: This technique is used to rewrite the general form of a circle equation into standard form.

Solving Quadratic Equations Using the Square Root Property

• Square Root Property: If $x^2 = a$, then $x = \pm \sqrt{a}$.

• Example: Solve $(x+2)^2 = 9$:

$\begin{align*} (x+2)^2 &= 9 \\ x+2 &= \pm 3 \\ x &= -2 \pm 3 \\ x &= 1 \text{ or } x = -5 \end{align*}$

• Example: Solve $(x+2)^2 = 7$:

$\begin{align*} (x+2)^2 &= 7 \\ x+2 &= \pm \sqrt{7} \\ x &= -2 \pm \sqrt{7} \end{align*}$

Introduction to Circles

A circle is the set of all points in a plane that are a fixed distance (radius) from a given point (center).

• Center: $(h, k)$

• Radius: $r$

• Standard Form of a Circle: $(x-h)^2 + (y-k)^2 = r^2$

Derivation Using the Distance Formula

• Distance Formula: The distance between $(x, y)$ and $(h, k)$ is $\sqrt{(x-h)^2 + (y-k)^2}$.

• Setting this distance equal to the radius $r$ gives the standard form:

$\sqrt{(x-h)^2 + (y-k)^2} = r$

Squaring both sides:

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

Writing the Equation of a Circle

To write the equation of a circle, you need the center $(h, k)$ and the radius $r$.

• General Formula: $(x-h)^2 + (y-k)^2 = r^2$

• Example: Center $(-2, 4)$, radius $6$:

$(x+2)^2 + (y-4)^2 = 36$

Finding the Equation of a Circle Given Center and a Point

If you know the center and a point on the circle, use the distance formula to find the radius.

• Example: Center $(0, 4)$, passes through $(4, 2)$:

$\begin{align*} r &= \sqrt{(4-0)^2 + (2-4)^2} \\ r &= \sqrt{16 + 4} \\ r &= \sqrt{20} \end{align*}$

Equation:

$(x-0)^2 + (y-4)^2 = 20$

Finding the Equation of a Circle Given Endpoints of a Diameter

When given endpoints of a diameter, first find the center (midpoint) and then the radius (half the distance between endpoints).

• Midpoint Formula: $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

• Example: Endpoints $(-4, 3)$ and $(2, -1)$:

$\begin{align*} \text{Center} &= \left( \frac{-4+2}{2}, \frac{3+(-1)}{2} \right) \\ &= (-1, 1) \end{align*}$

Radius is half the distance between endpoints:

$\begin{align*} d &= \sqrt{(2-(-4))^2 + ((-1)-3)^2} \\ &= \sqrt{36 + 16} \\ &= \sqrt{52} \\ r &= \frac{\sqrt{52}}{2} = \sqrt{13} \end{align*}$

Equation:

$(x+1)^2 + (y-1)^2 = 13$

Objective 2: Sketching the Graph of a Circle

To graph a circle, plot the center and use the radius to mark points in all directions from the center.

• Quadrant Location: The center's coordinates determine which quadrant the circle is located in.

• Example: Center $(2, 2)$, radius $2$:

$(x-2)^2 + (y-2)^2 = 4$

• Example: Center $(2, 4)$, radius $2$:

$(x-2)^2 + (y-4)^2 = 4$

Summary Table: Circle Equation Forms

Form	Description	Example
Standard Form	$(x-h)^2 + (y-k)^2 = r^2$	$(x+2)^2 + (y-4)^2 = 36$
General Form	$x^2 + y^2 + Dx + Ey + F = 0$	$x^2 + y^2 - 4x + 6y - 12 = 0$

Additional info: The general form can be converted to standard form by completing the square for both $x$ and $y$ terms.