# Circle theorems on the SAT test

### SAT Subscore: Additional topics in math

## Studying circle theorems

**On the SAT test circle theorems topic** is the fifth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

**Circle theorems topic is divided into sections** from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

**Finish studying** heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

### Circle theorems- summary

**Circle theorems topic includes two parts: **

Calculating arc length and sector areas using central angles measured in degrees.

Calculating angle measures in degrees inside a circle.

**The following formulas are provided at the beginning of each SAT math section:**

Circumference of a circle formula is C=2πr.

Area of a circle formula is A=πr^{2}.

Number of degrees of arc in a circle is 360.

**Pi (π)** is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal π (approximately 3.14).

**A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle** (one of its angles is a central angle and the other 2 angles are equal).

**The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii. **The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

**The relationship between central angle, arc length and sector area is given by the following ratios:**

central angle = arc length = sector area

______________ _____________________ ____________

360° circle circumference circle area

__Continue reading this page for detailed explanations and examples.__

### Calculating circumference and area of a circle

**The following formulas are provided at the beginning of each SAT math section:****Circumference of a circle** formula is C=2πr.**Area of a circle** formula is A=πr^{2}.

**Units of measurement:****Arc length and circumference** measure distance in units such as inches or centimeters.**Area and sector** measure area is square units such as square inches or square centimeters.**Angles** are measured in degrees.

Consider the following example:

The radius of the circle is equal to 10 centimeters.

What are the circumference and the area of the circle?

__Circumference of a circle:__

The circumference formula is C=2πr.

C=2*10*π=20π

__Area of a circle: __

The area formula is A=πr^{2}.

A=πr^{2}=π*10^{2}=100π

### The relationship between central angle, arc length and sector area

**A central angle **has a vertex at the center of a circle O and sides located on the circle circumference in two points A and B. The central angle ∠AOB determines a portion of the circumference (an arc) AC and a potion from the area (sector).

**A minor arc:** Each central angle divides a circle into two arcs. The smaller of the two arcs is called the minor arc and the larger of the two arcs is called the major arc (a minor arc subtends an angle less than or equal to 180°). __Note that__ when dealing with arcs, we always look at the minor arc.

**The relationship between central angle, arc length and sector area is given by the following ratios:**

central angle = arc length = sector area

______________ _____________________ ____________

360° circle circumference circle area

This relationship presents ** 3 equal ratios**:

Ratio 1: The **angle** as a portion of the total degrees of the circle (**360°**) **=**

Ratio 2: The **arc** length as a portion from the length of the circle circumference (**C=2****π****r**) **=**

Ratio 3: The **sector** area as a portion from the total area of the circle (**A=****π****r ^{2}**).

**Note that** we can calculate the denominators of the arc and the sector ratios given the value of the radius r.

**In the figure below we have the following ratios:**

∠AOC = arc AC = sector ACO area

_______ _______________ ________________

360° circumference circle area

** **

∠AOD = arc AD = sector ADO area

________ ________________ _________________

360° circumference circle area

∠BOD = arc BD = sector BDO area

________ _______________ __________________

360° circumference circle area

∠BOC = arc BC = sector BCO area

________ ______________ _________________

360° circumference circle area

Consider the following example:

A central angle of a circle is equal to 120 degrees and the radius of the circle is equal to 5 centimeters.

What is the arc length of the central angle?

__Calculating the circumference of the circle:__

C=2πr

C=2*5*π=10π

__Calculating the arc length with the ratios:__

Ratio 1: The **angle** as a portion of the total degrees of the circle (**360°**).

Ratio 2: The **arc** length as a portion from the length of the circle circumference (**C=2****π****r**).

central angle = arc length

______________ _____________________

360° circle circumference

**Representing with the variable x the arc length and plugging the given values into the ratio:**

120° = x

_____ _____

360° 10π

360x=1200π

x=1200/360π

x=120/36π

x=10/3π=3.33π

x=3^{1}/_{3}π=3^{1}/_{3}*3.14=10.47

The arc length is 10.47 centimeters.

__Checking the answer:__

120/360=1/3=0.33

3.33/10.47=0.32

The answers are different because of rounding differences.

Consider the following example:

The arc length of a circle is 2 centimeters and the radius of the circle is 1.6 centimeters.

What is the area of the sector of the arc?

__Calculating the circumference of the circle:__

C=2πr

C=2*1.6*π=3.2π=10.

__Calculating the area of the circle:__

The area formula is A=πr^{2}.

A=π*1.6^{2}=2.56π=8.

**Calculating the sector area with the ratios:**

Ratio 1: The **arc** length as a portion from the length of the circle circumference (**C=2****π****r**).

Ratio 2: The **sector** area as a portion from the total area of the circle (**A=****π****r ^{2}**).

**Representing with the variable x the sector area and plugging the given values into the ratio:**

arc length = sector area

____________________ _____________

circle circumference circle area

1.6 = x

____ ___

10 8

1.6*8=10x

x=1.6*8/10

x=1.28

The sector area is 1.28 centimeters.

__Checking the answer:__

1.6/10=0.16

1.28/8=0.16

### Angle relationships in circle

**This subject includes 2 angle types: **

Angles of isosceles triangles in a circle.

Angles created by intersection of diameters in a circle.

#### Angles of isosceles triangles in a circle

At the beginning of each SAT math section, it is provided that the number of degrees of arc in a circle is 360. In other words, **the sum of central angle measures in a circle is equal to 360°.**

**A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle.** Meaning that one of its angles is a central angle and the other 2 angles are equal.

**The sum of angles in a triangle is 180°**, if we are given the value of the central angle or the value of one of the equal angles, we can calculate the other angle values.

Consider the following example:

In the figure below, the point O is the center of a circle. The angle ∠AOB is equal to 100°.

What is the value of the angle ∠OAB?

Since the angle ∠AOB is a central angle, the sides OA and OB are radii of the circle and therefore the triangle ABO is an isosceles triangle.

In an isosceles triangle the angles opposite the two equal sides are equal, therefore ∠OAB=∠ABO.

The sum of angles in a triangle is equal to 180°, therefore ∠OAB+∠ABO+∠AOB=180°.

We are given that ∠AOB=100° and we know that ∠OAB=∠ABO, therefore 100°+∠OAB+∠ABO=180°.

Represent by x the equal angles ∠OAB and ∠ABO we get an equation that we can solve:

100+2x =180

2x=80

x=40

The value of the angle ∠OAB is 40 degrees.

__Checking the answer:__

40+40+100=180

180=180

Consider the following example:

In the figure below, the point O is the center of a circle.

The angle ∠AOB is twice bigger than the angle ∠COA.

The angle ∠COB is 5 times bigger than the angle ∠COA (the angle ∠COB is bigger than 180°).

What is the value of the angles ∠AOB and ∠COB?

Represent by x the angle ∠COA.

We are given that the angle ∠AOB is twice bigger than the angle ∠COA, therefore the angle ∠AOB=2x.

We are given that the angle ∠COB is 5 times bigger than the angle ∠COA, therefore the angle ∠COB=5x.

The sum of central angle measures in a circle is equal to 360° and we are given that the point O is the center of the circle, therefore we can write an equation and solve it:

x+2x+5x=360

8x=360

x=45

We know that the angle ∠AOB=2x, therefore the angle ∠AOB=90°.

We know that the angle ∠COB=5x, therefore the angle ∠COB=5*45=225 degrees.

#### Angles created by intersection of diameters in a circle

**The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii.**

The intersection creates 4 central angles: **2 pairs of equal vertical angles and 4 pairs of supplementary angles** (the sum of supplementary angles is 180°).

Given the value of one of the 4 central angles, we can calculate the other central angles.

In the figure below AB and CD are diameters of the circle.

What are the central and supplementary angles created by the diameters?

The intersection creates 4 equal radii AO, CO, DO and BO.

The intersection point O is at the center of the circle and it divides each diameter to 2 radii: t**he diameter AB is divided to radii AO and OB; the diameter CD is divided to radii CO and OD.**

**The intersection creates 4 central angles:** ∠COA, ∠AOD, ∠DOB and ∠BOC.

**The intersection creates 2 pairs of equal vertical angles:** ∠COA=∠BOD; ∠COB=∠AOD.

**The intersection creates 4 pairs of supplementary angles:** ∠COA+∠AOD=180°; ∠AOD+∠DOB=180°; ∠DOB+∠BOC=180°; ∠BOC+∠COA=180°.

Consider the following example:

In the figure below, O the is center of the circle and the chords AB and CD intersect at point O. The angle ∠BOD is equal to 40°.

What is the ratio between the angles ∠ACO and ∠ADO?

__Given the value of one of the 4 central angles, we can calculate the other central angles:__

We are given that ∠BOD=40°

∠AOC=∠BOD (vertical angles are equal), therefore, ∠AOC=40°.

∠BOD+∠AOD=180° (the sum of supplementary angles is 180°), therefore ∠AOD=180°-40°=140°.

__Calculating the angles in triangle ACO:__

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since CO=AO the triangle ACO is isosceles and the angles ∠ACO and ∠CAO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore ACO+CAO+AOC=180°.

**Plugging the values we get an equation that we can solve:**

x+x+40=180

2x=140

x=70

The angle ∠ACO is equal to 70 degrees.

__Calculating the angles in triangle ADO:__

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since AO=DO the triangle ADO is isosceles and the angles ∠DAO and ∠ADO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore DAO+ADO+AOD=180°.

We found that the angle ∠ACO is equal to 140 degrees.

**Plugging the values, we get an equation that we can solve:**

x+x+140=180

2x=40

x=20

The angle ∠ADO is equal to 20 degrees.

**The ratio between the angles ∠ACO and ∠ADO is** 70/20=7/2=3^{1}/_{2}

Note that we need to write the value of an angle ∠ACO in the nominator and not the opposite.

You just finished studying circle theorems topic, the fifth topic of additional topics in math!

Continue studying the next additional topic in math- angles, arc lengths and trig functions.