Circle

A circle is a two-dimensional shape where every point on the boundary is equidistant from a central point, called the center. The circle is actually the inner part of the circumference, i.e., the enclosed area inside the circle frame. This distance between the boundary and the center is called radius. The diameter is twice the radius, and it passes through the center, dividing the circle into two equal parts.

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

examples of circles with different circumferences.
More relevant components of the circle:
  • Radius: The distance from the center of the circle to any point on the circumference.
  • Diameter: A straight line passing through the center that connects two points on the circumference, equal to twice the radius.
  • Arc: A portion of the circumference.
  • Chord: A line segment connecting two points on the circle.
  • Tangent: A line that touches the circle at exactly one point.

Circumference

The perimeter or boundary length of the circle.
Can be calculated as: C=2πrC=2\pi r

Area

he space enclosed within the circle, calculated as A=πr2A = \pi r^2

Practice Circle

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

M is the center of the circle.

Perhaps AB=CD AB=CD

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Step-by-Step Solution

CD is a diameter, since it passes through the center of the circle, meaning it is the longest segment in the circle.

AB does not pass through the center of the circle and is not a diameter, therefore it is necessarily shorter.

Therefore:

ABCD AB\ne CD

Answer

No

Exercise #3

Which diagram shows a circle with a point marked in the circle and not on the circle?

Step-by-Step Solution

The interpretation of "in a circle" is inside the circle.

In diagrams (a) and (d) the point is on the circle, while in diagram (c) the point is outside of the circle.

Answer

Exercise #4

Which figure shows the radius of a circle?

Step-by-Step Solution

It is a straight line connecting the center of the circle to a point located on the circle itself.

Therefore, the diagram that fits the definition is c.

In diagram a, the line does not pass through the center, and in diagram b, it is a diameter.

Answer

Exercise #5

In which of the circles is the point marked in the circle and not on the circumference?

Video Solution

Step-by-Step Solution

Let's remember that the circular line draws the shape of the circle, and the inner part is called a disk.

Therefore, in diagram B, the point is located in the inner part, meaning inside the disk.

Answer

Exercise #6

Which diagram shows the radius of a circle?

Step-by-Step Solution

Let's remember that a radius is a line segment connecting the center of a circle to any point on the circle itself.

In drawing C we can see that the line coming from the center of the circle indeed connects to a point on the circle itself, while in the other drawings the lines don't touch any point on the circle.

Therefore, C is the correct drawing.

Answer

Exercise #7

In which of the circles is the segment drawn the radius?

Video Solution

Step-by-Step Solution

Let's remember that a radius is a line segment connecting the center of the circle to a point that lies on the circle itself.

In drawing A, the line doesn't touch any point on the circle itself.

In drawing B, the line doesn't pass through the center of the circle.

We can see that in drawing C, the line that extends from the center of the circle is indeed connected to a point on the circle itself.

Answer

Exercise #8

Where does a point need to be so that its distance from the center of the circle is the shortest?

Step-by-Step Solution

Let's remember that the circle is actually the inner part of the circumference, meaning the enclosed area within the frame of the circumference.

Therefore, a point whose distance is less than the radius from the center of the circle will necessarily be inside the circle.

Answer

Inside

Exercise #9

A point whose distance from the center of the circle is _______ than the radius, is outside the circle.

Step-by-Step Solution

Let's remember that the circle is actually the inner part of the circumference, meaning the enclosed area within the frame of the circumference.

Therefore, a point whose distance is greater than the center of the circle will necessarily be outside the circle.

Answer

greater

Exercise #10

Is it possible that the circumference of a circle is 8 meters and its diameter is 4 meters?

Video Solution

Step-by-Step Solution

To calculate, we will use the formula:

P2r=π \frac{P}{2r}=\pi

Pi is the ratio between the circumference of the circle and the diameter of the circle.

The diameter is equal to 2 radii.

Let's substitute the given data into the formula:

84=π \frac{8}{4}=\pi

2π 2\ne\pi

Therefore, this situation is not possible.

Answer

Impossible

Exercise #11

A circle has the following equation:
x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2

Point O is its center and is in the second quadrant (a0 a\neq0 )


Use the completing the square method to find the center of the circle and its radius in terms of 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 have 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, or in other words we will ensure that on the left side is 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 short 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 is8ax 8ax , which 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 short 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 is in 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)

Therefore, 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 #12

M is the center of the circle.

In the figure we observe 3 diameters?

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Answer

No

Exercise #13

Is there sufficient data to determine that

GH=AB GH=AB

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Answer

No

Exercise #14

M is the center of the circle.

Perhaps MF=MC MF=MC

MMMAAABBBCCCDDDEEEFFFGGGHHH

Video Solution

Answer

Yes

Exercise #15

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

Video Solution

Answer