A chord in a circle is a straight line that connects 2 2 any points that are on the circle.
• The chord passes inside the circle and not over it.
• The longest chord in the circle is the diameter
The radius is not a chord.

Chords in the circle

Practice Chords of a Circle

Examples with solutions for Chords of a Circle

Exercise #1

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 #2

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 #3

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 #4

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 #5

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 #6

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 #7

How many times longer is the radius of the red circle than the radius of the blue circle?

220

Video Solution

Answer

5

Exercise #8

How many times longer is the radius of the red circle, which has a diameter of 24, than the radius of the blue circle, which has a diameter of 12?

Video Solution

Answer

2

Exercise #9

How many times longer is the radius of the red circle than the radius of the blue circle?

168

Video Solution

Answer

2 2

Exercise #10

Calculate the length of the arc marked in red given that the circumference is 6.

50

Video Solution

Answer

56 \frac{5}{6}

Exercise #11

Calculate the length of the arc marked in red given that the circumference is 36.

20

Video Solution

Answer

2

Exercise #12

Calculate the length of the arc marked in red given that the circumference is 12.

60°60°60°

Video Solution

Answer

2

Exercise #13

Calculate the length of the arc marked in red given that the circumference is 12.

240

Video Solution

Answer

8

Exercise #14

Calculate the length of the arc marked in red given that the circumference is equal to 24.

150°150°150°

Video Solution

Answer

10 10

Exercise #15

Calculate the area of the section painted red given that the area of the circle is 12.

240

Video Solution

Answer

8