The part that is between $2$ points on the circle.

The arc is part of the circumference of the circle and does not pass inside the circle.

The part that is between $2$ points on the circle.

The arc is part of the circumference of the circle and does not pass inside the circle.

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

We are here to explain to you what an arc in a circle is in the easiest and most logical way.

First, let's remember what the shape of an arc is...

When you look up at the sky and see a rainbow, it looks like this, right?

**How about a hair tie? It looks quite similar as well:**

Now that we remember the shape of the arc, it will be easier for us to remember what an arc in a circle is.

An arc in a circle is the part between $2$ points on the circle.

Pay attention: the arc is on the circle and not inside it. It is part of the circumference of the circle and closely resembles the rainbows we see in everyday life.

**Let's show it in the figure:**

In front of us is a circle.

If we take $2$ points on top of the circle, for example, $A$ and $B$,

the part of the circle between these two points will be an arc.

Pay attention that we do not draw a line between the points inside the circle (a chord)

but rather we paint the top part of the circle as part of its circumference.

**Note:**

The arc can be of any length and even if it does not remind us of the arc we see in everyday life, it will still be an arc in a circle.

While it is on the circle between $2$ points as part of the circumference, the circle is called an arc.

**We will see examples where the arc in the circle does not look like an arc shape:**

**If you are interested in this article, you might also be interested in the following articles:**

- The center of the circle
- Circle
- Radius
- Diameter
- Pi
- The circumference perimeter
- Circular area
- Chords in a circle
- Central angle in a circle
- Perpendicular to the chord from the center of the circle
- Inscribed angle in a circle
- Distance from the chord to the center of the circle

**In the** **Tutorela** **blog, you will find a variety of articles about mathematics.**

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

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.

greater

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

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.

Inside

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

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.

_{A circle has the following equation:}_{$x^2-8ax+y^2+10ay=-5a^2$}_{Point O is its center and is in the second quadrant ($a\neq0$)}

_{Use the completing the square method to find the center of the circle and its radius in terms of $a$.}

**Let's recall** that ** the equation of a circle with its center at **$O(x_o,y_o)$

$(x-x_o)^2+(y-y_o)^2=R^2$Now, ** let's now have a look at the equation for the given circle**:

$x^2-8ax+y^2+10ay=-5a^2$

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:

L**et's recall **the __short formula for squaring a binomial:__

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

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

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 is**$8ax$, which has a negative sign):

$\underline{ x^2-8ax} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$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:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d
\end{cases}$ 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$(4</span><span class="katex">a)^2$, __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:

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

We'll do this **within the given equation**:

$x^2-8ax+y^2+10ay=-5a^2 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4a}\underline{\underline{+\textcolor{green}{(4a)}^2-(4a)^2}}+y^2+10ay=-5a^2\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2+y^2+10ay=-5a^2\\$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 **$10ay$ has a positive sign)

$(x-4a)^2-16a^2+\underline{y^2+10ay}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a}=-5a^2\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a\underline{\underline{+(5a)^2-(5a)^2}}}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{5a}+\textcolor{green}{(5a)}^2-25a^2}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+(\textcolor{red}{y}+\textcolor{green}{5a})^2-25a^2=-5a^2\\ \boxed{(x-4a)^2+(y+5a)^2=36a^2}$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:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4a})^2+(y+\textcolor{orange}{5a})^2=\underline{\underline{36a^2}}\\ \downarrow\\ (x-\textcolor{purple}{4a})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{5a}))^2=\underline{\underline{36a^2}}\\$

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:**$\boxed{O(x_o,y_o)\leftrightarrow O(4a,-5a)}$ and extract the **radius of the circle** by solving a simple equation:

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

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,

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:

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

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

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

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

Test your knowledge

Question 1

Which diagram shows the radius of a circle?

Question 2

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

Question 3

Is it correct to say 'the area of a circle'?