The circle is actually the inner part of the circumference, i.e., the enclosed area inside the circle frame.

Below are some examples of circles with different circumferences. The colored part in each represents the circle:

examples of circles with different circumferences.

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Test yourself on parts of the circle!

einstein

There are only 4 radii in a circle.

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The circle is the inner part, which is colored (green, blue, orange). The circumference is just the outline, here colored black.

Often, as you progress in your studies, you will have to calculate the area of the circle or the perimeter of the circumference (in the following articles we will see how this is done). The area of the circle is the region that is bounded by the circumference (by the contour). The perimeter of the circumference is the length of the contour of the circle.

When we talk about area we should say area of the circle and not area of the circumference, although it is true that sometimes it is used by mistake and, therefore, you may come across the expression "area of the circumference".

Example:

B - dots on the perimeter of different circles

We have drawn a red dot for each of the illustrations. In the illustration on the right the dot is inside the area of the circle. We can also say that it is inside the perimeter of the circle.

In the middle illustration the dot is outside the area of the circle. We can also say that it is outside the perimeter of the circle. In the left illustration the red dot is on the perimeter of the circle.


Other key terms: center, radius, chord and diameter

Center

The center is the interior point equidistant to all points of the perimeter. Usually this point is marked with the letter O.

In these illustrations the center of the circle is marked with a black dot:

The center of the circumference


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Radio

The radius is the distance between the center of the circle and any other point on the perimeter. It is denoted by the capital letter R or lowercase r as follows:

P6 - Radius

We will see that, intuitively, the larger the area of the circle and perimeter, the larger the length of the radius. Next we will learn more peculiarities of the relationship between them.


Chord

A chord is a straight line joining two points on the perimeter of the circle. We can draw an infinite number of chords on any circle. Note that the chord does not necessarily have to go through the center of the circle. For example, look at the chords in the illustration below:

Chords in the circle


Do you know what the answer is?

Diameter

The diameter of the circle is the chord that passes exactly through the center. That is, it is the straight line joining two points of the perimeter passing through the center of the circle. It is usually denoted by the letter D. It looks like this:

p4- Diameter


Circle Exercises

Exercise 1

Assignment:

Given the circumference of the figure

The diameter of the circle is 13 13 ,

What is its area?

Exercise 1 Assignment Given the circumference of the figure

Solution

It is known that the diameter of the circle is twice its radius, i.e. it is possible to know the radius of the circle in the figure.

13:2=6.5 13:2=6.5

To find the area of the circle, we replace the data we have in the formula for the calculation of the circle

A=π×R2 A=\pi\times R²

We replace the data we have:

A=π×6.52 A=\pi\times6.5²

A=π42.25 A=\pi42.25

Answer

42.25π 42.25\pi


Check your understanding

Exercise 2

Request

Given an equilateral triangle in a circle

What is the area of the circle?

Exercise 2 Assignment Given an equilateral triangle in a circle

Solution

Recall first the theorem that a circumferential angle that is inclined about the diameter is equal to 90o 90^o degrees.

That is, the triangle inside the circle is a right triangle and isosceles, so we can use the Pythagorean formula.

We replace the data we have with the Pythagorean formula

X=Diaˊmetro X=Diámetro

(2)2+(2)2=X2 (\sqrt{2})²+(\sqrt{2})²=X²

The root cancels the power and therefore we obtain that

2+2=X2 2+2=X²

That is

X2=4 X²=4

The root of 44 is 22 and therefore the diameter is equal to 22 and the radius is equal to half of the diameter and therefore it is equal to 11

Therefore we obtain that the diameter is equal to 22

And the radius is 1 1

Then we add the formula for the area of the circle

Answer

π π


Exercise 3

Given the semicircle:

new Exercise 3- Given the semi-circle

Consigna

Calculate its area

Solution

Since we know that it is a semicircle we can conclude that the base of the semicircle is the diameter.

We know that the diameter is twice the radius and therefore we can know the radius of the circle.

Diaˊmetro=14 Diámetro = 14

14:2=7 14:2=7

Radio=7 Radio = 7

The formula for calculating the area of the circumference is

A=π×R2 A=\pi\times R²

We replace the data in the formula

A=π×72 A=\pi\times 7²

A=π×49 A=\pi\times 49

Since the formula was the area of the semicircle we divide the area of the circle by 2 2 and get the answer.

π49:2=24.5π \pi49:2=24.5\pi

Answer

π49:2=24.5π \pi49:2=24.5\pi


Do you think you will be able to solve it?

Exercise 4

Query

Given a circle whose circumference 6.28 6.28

What is the area?

Exercise 4- Assignment Given a circle whose circumference 6.28

Solution

To solve this question we will use two formulas:

The first formula is to calculate the circumference:

P=2×π×R P=2×π×R

3.14=pi×R 3.14=pi\times R

We divide by (3.14) (3.14)

R=1 R=1

The second formula is for calculating the area of the circle

A=π×R×R A=π×R×R

We replace the answer we got

A=π×1×1 A=π×1×1

A=π A=π

Answer

π π


Exercise 5

Request

Given the shape of the figure

A quadrilateral is a square with the length of its sides 5cm 5\operatorname{cm}

For each side extends a semicircle.

What is the circumference of this shape?

Given the shape of the figure A quadrilateral is a square whose length of the sides

Solution

The circumference consists of 4 4 semicircles.

412p=2p 4\cdot\frac{1}{2}p=2p

That is, in total 2 2 circumference than its diameter 5cm 5\operatorname{cm}

Diameter = Radius multiplied by 2 2

The radius multiplied by 2=5 2=5

Divide by 2 2

radio=2.5cm radio=2.5 cm

p=2π2.5=5π p=2\pi\cdot2.5=5\pi

We calculate the area of the form

2p=25π=10π 2\cdot p=2\cdot5\pi=10\pi

Answer

10π 10\pi


Test your knowledge

Exercise 6

Request

Given the shape of the figure

For the sides of the triangle two semicircles are extended.

The triangle is equilateral and each side has a length of 6Xcm 6X\operatorname{cm}

What is the circumference of the shape?

Exercise 6 sides of the triangle extend two semicircles

Solution

P=ladotriaˊngulo+2(12P) P=ladotriángulo+2\cdot\left(\frac{1}{2}P\right)

Diameter of the circle

= 6Xcm 6X\operatorname{cm}

Each semicircle contributes to the circumference of the semicircle of the whole circle the diameter of one side of a triangle.

p=6x+2(122πr)= p=6x+2\cdot(\frac{1}{2}\cdot2\pi r)=

We reduce by : 2 2

6x+2π(6x2)= 6x+2\cdot\pi(\frac{6x}{2})=

We reduce by: 2 2

6x+6πx 6x+6\pi x

Answer

6x+6πx 6x+6\pi x


Exercise 7

Request

Given the circle in the figure

The radius of the circle is equal to: 9.5 9.5

What is its circumference?

Exercise 7 -Given radius of circle is equal to 9.5

Solution

The radius of the circle is r=912 r=9\frac{1}{2}

We use the formula of the circumference

2πr 2\pi r

We replace accordingly and get

2π912=19π 2\cdot\pi\cdot9\frac{1}{2}=19\pi

Answer

19π 19\pi


Do you know what the answer is?

Exercise 8

Consigna

A construction company offered two tents for the kindergarten.

The circles are identical in each tent and form holes.

Which tent will generate more shade?

Exercise 7 -Question A construction company offered two tents for the kindergarten.

Solution

The shade depends on the area of the tent:

S1=784π(22)2= S_1=7\cdot8-4\cdot\pi(\frac{2}{2})^2=

564π(22)2=43.44 56-4\pi(\frac{2}{2})^2=43.44

S2=9203π(32)2= S_2=9\cdot20-3\cdot\pi(\frac{3}{2})^2=

9203π(32)2= 9\cdot20-3\pi(\frac{3}{2})^2=

9206.75π=158.805 9\cdot20-6.75\pi=158.805

Answer

B B


Review questions

What is a circle?

A circle is that part that is enclosed in a curved line called a circumference, in the following image the circle is the part that is blue.

What is a circle


Check your understanding

What is the difference between circle and circumference?

The circle is the part that is inside the circumference, and the circumference is the line that surrounds the circle, let's see the difference with the following image.

What is the difference between circle and circumference


What are the parts of a circle?

A circle is made up of a center, a radius, a diameter and a circumference.

What are the parts of a circle


Do you think you will be able to solve it?

What is a unit circle?

That which has a radius equal to 1 1 is called a unit circle.

A circle with a radius equal to 1 is called a unit circle.


What is the area of a circle and how is it calculated?

The Area of a circle is the surface, that is, the inner part of the whole circumference and we can calculate it with the formula A=πR2 A=\pi R^2

Let's see an example:

Assignment:

Calculate the area of the following circle with D=9 cm D=9\text{ cm}

Calculate the area of the following circle with D=9 cm

We know that the radius is half the diameter, therefore:

R=4.5 cm R=4.5\text{ cm}

π=3.14 \pi=3.14

Now we substitute this data into the area formula:

A=πR2=3.14(4.5 cm)2 A=\pi R^2=3.14\left(4.5\text{ cm} \right)^2

3.14(20.25cm2)=63.585cm2 3.14\left(20.25\operatorname{cm}^2 \right)=63.585\operatorname{cm}^2

Answer

A=63.585cm2 A=63.585\operatorname{cm}^2


Test your knowledge

examples with solutions for circle

Exercise #1

There are only 4 radii in a circle.

Step-by-Step Solution

A radius is a straight line that connects the center of the circle with a point on the circle itself.

Therefore, the answer is incorrect, as there are infinite radii.

Answer

False

Exercise #2

Given the circle with the equation:

x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2


and its center at point O in the second quadrant,

a0 a\neq0

Use the completing the square method to find the center of the circle and its radius using

a a

Step-by-Step Solution

 Let's recall that the equation of a circle with its center at O(xo,yo) O(x_o,y_o) and its radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2 Now, let's now take a look at the equation for the given circle:

x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2 We will try rearrange this equation to match the circle equation, that is - we will ensure that on the left side there will be the sum of two squared binomial expressions, one for x and one for y.

We will do this using the "completing the square" method:

Let's recall the shortcut formula for squaring a binomial:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2 We'll deal separately with the part of the equation related to x in the equation (underlined):

x28ax+y2+10ay=5a2 \underline{ x^2-8ax}+y^2+10ay=-5a^2

We'll isolate these two terms from the equation and deal with them separately.

We'll present these terms in a form similar to the form of the first two terms in the shortcut formula (we'll choose the subtraction form of the binomial squared formula since the term in the first power we are dealing with 8ax 8ax has a negative sign):

x28axc22cd+d2x22x4ac22cd+d2 \underline{ x^2-8ax} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\ Notice that compared to the shortcut formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the comparison:

{xc4ad \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d \end{cases} Therefore, if we want to get a squared binomial form from these two terms (underlined in the calculation), we will need to add the term(4</span><spanclass="katex">a)2 (4</span><span class="katex">a)^2 , but we don't want to change the value of the expression, and therefore - we will also subtract this term from the expression.

That is, we will add and subtract the term (or expression) we need to "complete" to the binomial squared form,

In the following calculation, the "trick" is highlighted (two lines under the term we added and subtracted from the expression),

Next - we'll put the expression in the squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression:

x22x4ax22x4a+(4a)2(4a)2x22x4a+(4a)216a2(x4a)216a2 x^2-2\cdot x\cdot 4a\\ x^2-2\cdot x\cdot4a\underline{\underline{+(4a)^2-(4a)^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}+(\textcolor{green}{4a})^2-16a^2\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2}\\ Let's summarize the steps we've taken so far for the expression with x.

We'll do this within the given equation:

x28ax+y2+10ay=5a2x22x4a+(4a)2(4a)2+y2+10ay=5a2(x4a)216a2+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4a}\underline{\underline{+\textcolor{green}{(4a)}^2-(4a)^2}}+y^2+10ay=-5a^2\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2+y^2+10ay=-5a^2\\ We'll continue and do the same thing for the expressions with y in the resulting equation:

(Now we'll choose the addition form of the squared binomial formula since the term in the first power we are dealing with 10ay 10ay has a positive sign)

(x4a)216a2+y2+10ay=5a2(x4a)216a2+y2+2y5a=5a2(x4a)216a2+y2+2y5a+(5a)2(5a)2=5a2(x4a)216a2+y2+2y5a+(5a)225a2=5a2(x4a)216a2+(y+5a)225a2=5a2(x4a)2+(y+5a)2=36a2 (x-4a)^2-16a^2+\underline{y^2+10ay}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a}=-5a^2\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a\underline{\underline{+(5a)^2-(5a)^2}}}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{5a}+\textcolor{green}{(5a)}^2-25a^2}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+(\textcolor{red}{y}+\textcolor{green}{5a})^2-25a^2=-5a^2\\ \boxed{(x-4a)^2+(y+5a)^2=36a^2} In the last step, we move the free numbers to the second side and combine like terms.

Now that the given circle equation isin the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius:

(xxo)2+(yyo)2=R2(x4a)2+(y+5a)2=36a2(x4a)2+(y(5a))2=36a2 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4a})^2+(y+\textcolor{orange}{5a})^2=\underline{\underline{36a^2}}\\ \downarrow\\ (x-\textcolor{purple}{4a})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{5a}))^2=\underline{\underline{36a^2}}\\ In the last step, we made sure to get the exact form of the general circle equation - that is, where only subtraction is performed within the squared expressions (emphasized with an arrow)

Therefore, we can conclude that the center of the circle is at:O(xo,yo)O(4a,5a) \boxed{O(x_o,y_o)\leftrightarrow O(4a,-5a)} and extract the radius of the circle by solving a simple equation:

R2=36a2/R=±6a R^2=36a^2\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\pm6a}

Remember that the radius of the circle, by its definition is the distance between any point on the diameter and the center of the circle. Since it is positive, we must disqualify one of the options we got for the radius.

To do this, we will use the remaining information we haven't used yet - which is that the center of the given circle O is in the second quadrant,

That is:

O(x_o,y_o)\leftrightarrow x_o<0,\hspace{4pt}y_o>0 (Or in words: the x-value of the circle's center is negative and the y-value of the circle's center is positive)

That is, it must be true that:

\begin{cases} x_o<0\rightarrow (x_o=4a)\rightarrow 4a<0\rightarrow\boxed{a<0}\\ y_o>0\rightarrow (y_o=-5a)\rightarrow -5a>0\rightarrow\boxed{a<0} \end{cases} We concluded that a<0 and since the radius of the circle is positive we conclude that necessarily:

R=6a \rightarrow \boxed{R=-6a} Let's summarize:

O(4a,5a),R=6a \boxed{O(4a,-5a), \hspace{4pt}R=-6a} Therefore, the correct answer is answer d. 

Answer

O(4a,5a),R=6a O(4a,-5a),\hspace{4pt}R=-6a

Exercise #3

Is there sufficient data to determine that

GH=AB GH=AB

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Answer

No

Exercise #4

M is the center of the circle.

Perhaps AB=CD AB=CD

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Answer

No

Exercise #5

In which of the circles is the center of the circle marked?

Video Solution

Answer

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