# Central Angle in a Circle

🏆Practice the parts of a circle

### Central angle in a circle

We are here to define what a central angle in a circle is and give you tips to remember its definition and properties in the best and most logical way.
Before talking about the central angle in a circle, let's take a moment to look at its name - a central angle.

Through its name, we can recognize that it has some connection with the center of the circle.
Great, now let's move on to the definition of a central angle and it will make much more sense to us.

### What is a central angle in a circle?

A central angle in a circle is an angle whose vertex is the center of the circle and its ends are the radii of the circle
Therefore, its ends are on the top part of the circle.
If we connect all the central angles in the same complete circle - we will obtain $360°$.

## Test yourself on the parts of a circle!

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

### In front of us, there is a circle.

We will mark the center of the circle with the letter $A$.

We note that a central angle is an angle whose vertex is the center of the circle
and its sides are the radii of the circle.
Therefore, if we draw two radii, an angle will be formed.
The angle that will be created will be a central angle in the circle.

Now that we already know what a central angle in a circle is and can easily identify it,
we must learn about some important theorems and characteristics of a central angle in a circle.
Shall we begin?

## Central angle in the circle

When can we determine that two central angles in a circle are equal?
In two cases:

• If the arcs on which the angles are based are equal, then we can determine that the central angles are equal.

Let's see this in the figure:

If $BC=DE$
Then $∢A1=∢A2$

If the arcs in front of the central angles are equal, then the central angles are equal.

In the same way, the theorem works in reverse.
If the central angles are equal, the arcs in front of them are also equal.

• If the chords opposite the central angles are equal, then we can determine that the central angles are equal.

Let's see this in the figure:

If $BC=DE$
then $∢A1=∢A2$

If the chords in front of the central angles are equal, then the central angles are equal.
In the same way, the theorem works in reverse.
If the central angles are equal, the chords in front of them are also equal.

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## Inscribed angle

Now we will learn the relationship between a central angle in a circle and an inscribed angle in a circle.
Remember that an inscribed angle in a circle is an angle whose vertex is on the circumference of the circle and whose ends are chords in a circle.
Like here:

## The relationship between a central angle and an inscribed angle in a circle

In a circle, the central angle will be twice the inscribed angle that spans the same arc.
Or, the inscribed angle will be equal to half the central angle that spans the same arc.

That is:
If in the circle we identify a central angle and an inscribed angle that lean on the same arc, we can say that as they lean on the same arc, the inscribed angle will be equal to half of the central angle or alternatively, the central angle will be double the inscribed angle.
And we will see it in the illustration:

Before us, there is a circle.

We can identify that the angle $A$ is a central angle, which comes from the vertex of the circle and its radii at the ends, while an angle $B$ is an inscribed angle - a vertex is on the circle and its ends are chords.
We can also see that these two angles are subtended - they are on the same arc - $CD$
Therefore, we can conclude that the inscribed angle $B$ is equal to half of the central angle $A$ or alternatively, the angle $A$ is equal to twice the angle $B$.

$∢A=α$
$∢B=1/2 α$

or

$∢A=2α$
$∢B=α$

Wonderful!

Now we understand the relationship between an inscribed angle and a central angle. It is worth knowing more different representations of inscribed and central angles that are on the same arc:

## Examples and exercises with solutions of let's see a numerical example

### Exercise #1

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 $a$.

### Step-by-Step Solution

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

$(x-x_o)^2+(y-y_o)^2=R^2$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:

Let'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$(4a)^2$, 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, 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:

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

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

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

### Exercise #2

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

### Exercise #3

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

### Exercise #4

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

5

### Exercise #5

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

2

If in a circle we are given that the inscribed angle is $5$ degrees and we are asked what the value of the marked central angle is,
we can see that they lean on the same arc and therefore determine that the central angle is $2$ times greater than the inscribed angle that leans on the same arc.
It is deduced
that the marked central angle is equal to $10$.
$5\times2=10$

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