Chapter 1: Triangles and Circles

1.3 Circles

Algebra Refresher

Algebra Refresher for Circles

True or False?

1. [latex]\sqrt{a^{2}+b^{2}}=a+b[/latex]
2. [latex]\sqrt{36+64}=6+8[/latex]
3. [latex]\sqrt{16x^{4}}=4x^{2}[/latex]
4. [latex]\sqrt{2x}\sqrt{3y}=\sqrt{6xy}[/latex]
5. [latex]\sqrt{5x}+\sqrt{3x}=\sqrt{8x}[/latex]
6. [latex]\sqrt{4+N}=2+\sqrt{N}[/latex]
7. [latex]\sqrt{\frac{x}{4}}=\frac{\sqrt{x}}{2}[/latex]
8. [latex]\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}[/latex]

 

Algebra Refresher Answers
  1. False
  2. False
  3. True
  4. True
  5. False
  6. False
  7. True
  8. True

 

Circles

Learning Objectives

  • Find the distance between two points
  • Distinguish between exact values and approximations
  • Graph a circle
  • Find and use the equation for a circle
  • Find the length of a fraction of a circle
  • Find the area of a sector of a circle

 

The Distance Formula

Delbert is hiking in the Santa Monica Mountains, and he would like to know the distance from the Sycamore Canyon trailhead, located at [latex]12[/latex]-C on his map, to the Coyote Trail junction, located at [latex]8[/latex]-F, as shown below.

SM mountains

Each interval on the map represents one kilometer. Delbert remembers the Pythagorean theorem and uses the map coordinates to label the sides of a right triangle. The distance he wants is the hypotenuse of the triangle, so
[latex]d^{2} = 4^{2} + 3^{2} = 16 + 9 =25 \\ d = \sqrt{25} = 5[/latex]

The straight-line distance to Coyote junction is about [latex]5[/latex] kilometers.
The formula for the distance between two points is obtained in the same way.

We first label a right triangle with points [latex]P_{1}[/latex] and [latex]P_{2}[/latex] on opposite ends of the hypotenuse. (See the figure at right.) The sides of the triangle have lengths [latex]\lvert{x_2-x_1}\rvert[/latex] and [latex]\lvert{y_2-y_1}\rvert[/latex]. We can use the Pythagorean theorem to calculate the distance between [latex]P_{1}[/latex] and [latex]P_{2}[/latex]
[latex]d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}[/latex]

distance formula

Taking the (positive) square root of each side of this equation gives us the distance formula.

Distance Formula

The distance [latex]d[/latex] between two points [latex]P_{1}(x_{1}, y_{1})[/latex] and [latex]P_{2}(x_{2}, y_{2})[/latex] is
[latex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/latex]

 

Example 1.3.1.

Find the distance between [latex](2,-1)[/latex] and [latex](4,3)[/latex]

Solution.

We substitute [latex](2,-1)[/latex] for ([latex]x_{1}[/latex], [latex]y_{1}[/latex]) and [latex](4,3)[/latex] for ([latex]x_{2}[/latex], [latex]y_{2}[/latex]) in the distance formula to obtain
[latex]d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \\ = \sqrt{(4-2)^{2}+[3-(-1)]^{2}} \\ = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}[/latex]
The exact value of the distance, shown at right, is [latex]2\sqrt{5}[/latex] units. We obtain the same answer if we use [latex](4,3)[/latex]for [latex]P_{1}[/latex]and [latex](2,-1)[/latex] for [latex]P_{2}[/latex]
[latex]d = \sqrt{(2-4)^{2}+[(-1)-3]^{2}} \\ = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}[/latex]

We can use a calculator to obtain an approximation for this value, and find
[latex]  2\sqrt{5}\approx 2(2.236)=4.472[/latex]

distance formula

 

 

Caution 1.32.

In the previous example, the radical [latex]\sqrt{4+16}[/latex] cannot be simplified to [latex]\sqrt{4}+\sqrt{16}[/latex]. (Do you remember why not?)

 

Checkpoint 1.33.

  1. Find the distance between the points [latex](-5,3)[/latex] and [latex](3,-9)[/latex].
  2. Plot the points on a Cartesian grid and show how the Pythagorean theorem is used to calculate the distance.
Answer.

[latex]4\sqrt{13}[/latex]

Equation for a Circle

A circle is the set of all points in a plane that lie at a given distance, called the radius, from a fixed point called the center. We can use the distance formula to find an equation for a circle.
The circle shown below has its center at the origin, [latex](0,0)[/latex] and its radius is [latex]r[/latex].

Now, the distance from the origin to any point [latex]P(x,y)[/latex] on the circle is [latex]r[/latex]. Therefore,
[latex]\sqrt{(x-0)^{2}+(y-0)^{2}} = r[/latex]
or, squaring both sides,
[latex](x-0)^{2}+(y-0)^{2}=r^2[/latex]

circle

Because every point on the circle must satisfy this equation, we have found an equation for the circle.

Circle

The equation for a circle of radius [latex]r[/latex] centered at the origin is
[latex]x^{2}+y^{2}=r^{2}[/latex]

 

Example 1.34.

Find two points on the circle [latex]x^{2}+y^{2}=4[/latex] with x-coordinate [latex]-1[/latex].

Solution.

We substitute [latex]x=-1[/latex] into the equation for the circle, and solve for [latex]y[/latex].

[latex](-1)^{2}+y^{2} = 4 \\ y^{2} = 4-1 = 3 \\ y = \pm \sqrt{3}[/latex]
The points are [latex](-1, \sqrt{3})[/latex] and [latex](-1,-\sqrt{3})[/latex] as shown at right. Note that [latex]\sqrt{3} \approx 1.732[/latex].

circle r=2

 

Checkpoint 1.35.

Find the coordinates of two points on the circle [latex]x^{2}+y^{2}=1[/latex] with [latex]y[/latex]-coordinate [latex]\frac{1}{2}[/latex]

Answer.

[latex]\bigg( \frac{\sqrt{3}}{2},\frac{1}{2}\bigg)[/latex],  [latex]\bigg( \frac{-\sqrt{3}}{2},\frac{1}{2} \bigg)[/latex]

 

Definition 1.36. Unit Circle

The circle in the exercise above, [latex]x^{2}+y^{2}=1[/latex], which is centered at the origin and has radius [latex]1[/latex] unit, is called the unit circle.

 

Rational and Irrational Numbers

Every common fraction, such as [latex]\frac{3}{4}[/latex], can be written in many equivalent forms, including a decimal form. For example,
[latex]\frac{3}{4}=\frac{6}{8}=\frac{75}{100}=0.75 ~~~and~~~\frac{5}{11}=\frac{20}{44}=\frac{50}{110}=0.\overline{45}[/latex]

where [latex]0.\overline{45}[/latex] is the repeating decimal [latex]0.45454545....[/latex]
Because the fraction bar denotes division, a fraction is a quotient of two integers, and we can calculate its decimal form by dividing the denominator into the numerator.

Definition 1.37. Rational Number

Any number (including fractions) that can be written as a quotient of two integers [latex]\frac{a}{b},~~\text{where}~~ b\neq 0[/latex] is called a rational number.

 

The decimal form of a rational number is either a terminating decimal, such as [latex]0.75[/latex] or a repeating decimal, such as [latex]0.\overline{45}[/latex]. Thus, we can always write down an exact decimal equivalent for a rational number, although we may choose to round off a particularly long or unwieldy decimal. For example,
[latex]\frac{3}{7}=0.\overline{428571}\approx 0.43[/latex]

Caution 1.38.

Note that [latex]0.43[/latex] is not exactly equal to [latex]\frac{3}{7}[/latex]; it is an approximation for [latex]\frac{3}{7}[/latex] just as [latex]0.33[/latex] is an approximation for [latex]\frac{1}{3}[/latex]. In our work, it will be important to distinguish between exact values and approximations.

 

Definition 1.39. Irrational Number.

An irrational number is one that cannot be written as a quotient of two integers [latex]\frac{a}{b}[/latex] where [latex]b \neq 0[/latex].

 

Examples of irrational numbers are [latex]\sqrt{3}[/latex], [latex]\sqrt{5}[/latex], and [latex]\pi[/latex]. The decimal form of an irrational number is nonterminating and nonrepeating. The first few digits of the examples mentioned are
[latex]\sqrt{3}  = 1.732050807568...\\ \sqrt{5}  = 2.236067977..\\ \pi  = 3.141592653589...[/latex]

but none of these decimal forms ever ends. Thus, we cannot write down an exact decimal equivalent for an irrational number. The best we can do is give a decimal approximation, no matter how many digits we include.

Example 1.40.

Which values are exact, and which are approximations?

  1. [latex]\frac{7}{32}\rightarrow 0.21875[/latex]
  2. [latex]\sqrt{8}\rightarrow 2.828427125[/latex]
  3. [latex]\sqrt{0.16}\rightarrow 0.4[/latex]
  4. [latex]\frac{\pi}{2}\rightarrow 1.570796327[/latex]
Solution.

a. Because [latex]\frac{7}{32}[/latex] is a rational number, it has an exact decimal equivalent. Divide [latex]7[/latex] by [latex]32[/latex] to see that [latex]\frac{7}{32}=0.21875[/latex].

b. Because [latex]8[/latex] is not a perfect square, ([latex]\sqrt{8}[/latex]) is irrational, so [latex]2.828427125[/latex] is not the exact value of ([latex]\sqrt{8}[/latex]).

c. [latex]\sqrt{0.16}=\sqrt{\frac{16}{100}}=\frac{4}{10}=0.4[/latex], so this value is exact.

d. Because ([latex]\frac{\pi}{2}[/latex]) is an irrational number, it has no decimal equivalent, so (1.570796327) is an approximation.

 

Checkpoint 1.41.

For which numbers can you give an exact decimal equivalent?

  1. [latex]\frac{\sqrt{3}}{2}[/latex]
  2. [latex]\frac{\sqrt{16}}{3}[/latex]
  3. [latex]2 \pi[/latex]
  4. [latex]\frac{25}{17}[/latex]
Answer.

b, d

 

 

Circumference and Area

Recall from geometry that the circumference of a circle is proportional to its radius.

Circumference of a Circle

The circumference of a circle of radius [latex]r[/latex] is given by

[latex]C=2 \pi r[/latex]

 

The number [latex]\pi[/latex] gives the ratio of the circumference of any circle to its diameter. It is an irrational number, [latex]\pi \approx 3.14159[/latex].
The length of a portion, or arc, of a circle is called its arclength.

Example 1.42.

Francine baked an apple pie with diameter [latex]8[/latex] inches. If Delbert cuts himself a [latex]60°[/latex] wedge, what is the arclength of the curved edge?
Solution.

The radius of the pie is [latex]4[/latex] inches, so its circumference is [latex]2 \pi 4[/latex] inches.

A [latex]60°[/latex] wedge, shown at right, is [latex]\frac{1}{6}[/latex] of the entire pie, so its edge is [latex]\frac{1}{6}[/latex] of the circumference. The exact length of the arc is thus
[latex]\frac{1}{6} (8 \pi) = \frac{4 \pi}{3} \text{ inches.}[/latex]
Using a calculator, we find that [latex]\frac{4 \pi}{3}=4.19[/latex] rounded to two decimal places, so the length of the curved edge is between [latex]4[/latex] and [latex]4\bigg( \frac{1}{2} \bigg)[/latex] inches.

pie

 

Checkpoint 1.43.

What is the arclength of the curved edge of a [latex]60°[/latex] wedge cut from a blueberry pie of diameter [latex]10[/latex] inches?

Answer.

[latex]\frac{5 \pi}{3} \approx 5.24[/latex] inches.

 

The area of a circle is proportional to the square of its radius.

Area of a Circle

The area of a circle of radius [latex]r[/latex] is given by
[latex]A=\pi r^{2}[/latex]

 

A portion of a circle shaped like a pie-shaped wedge is called a sector.

Example 1.44.

What is the area of Delbert’s slice of pie in the previous example?

Solution. As we saw in the previous example, Delbert’s sector of the pie is [latex]\frac{1}{6}[/latex] of the entire pie, so its area [latex]\frac{1}{6}[/latex] of the area of the whole pie, or
[latex]\frac{1}{6} \pi (4^{2})= \frac{8 \pi}{3}~~ \text{square inches}[/latex]. The area of the wedge is [latex]\frac{8\pi}{3}[/latex], or about [latex]8.34[/latex] square inches.

 

Checkpoint 1.45.

What is the area of a [latex]60°[/latex] wedge cut from a blueberry pie of diameter [latex]10[/latex] inches?

Answer.

[latex]\frac{25 \pi}{6} \approx 13.09 \ \text{square inches}[/latex]

 

Section 1.3 Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Concepts

Distance Formula.
  1. The distance (d) between two points ([latex]P_{1}(x_{1}, y_{1})[/latex]) and ([latex]P_{2}(x_{2}, y_{2})[/latex]) is
    [latex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/latex]
  2. Any number that can be written as a quotient of two integers ([latex]~~\frac{a}{b},~~\text{where}~~ b\neq 0[/latex]) is called a rational number. The decimal form of a rational number is either a terminating decimal or a repeating decimal.
  3. An irrational number is one that cannot be written as a quotient of two integers ([latex]~~\frac{a}{b},~~\text{where}~~ b \neq 0[/latex]). We cannot write down an exact decimal equivalent for an irrational number.
  4. A circle is the set of all points in a plane that lie at a given distance, called the radius, from a fixed point called the center.

Circle.

  1. The equation for a circle of radius (r) centered at the origin is [latex]x^{2}+y^{2}=r^{2}[/latex]
  2. The circle ([latex]~x^{2}+y^{2}=1[/latex]), which is centered at the origin and has radius 1 unit, is called the unit circle.

Circumference of a Circle.

  1. The circumference of a circle of radius (r) is given by [latex]C=2\pi r[/latex]

Area of a Circle.

  1. The area of a circle of radius (r) is given by [latex]A=\pi r^{2}[/latex]

 

Study Questions

  1. Explain why the distance formula, ([latex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/latex]), cannot be simplified to ([latex](x_{2}-x_{1})+(y_{2}-y_{1})[/latex]).
  2. What is a unit circle, and what is its equation?
  3. Can you give an exact decimal value for [latex]\pi\text{?}[/latex]
  4. What is the arclength of a semicircle of radius [latex]r[/latex]?
definition

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