Inscribed angle in a circle

🏆Practice the parts of a circle

An inscribed angle in a circle is an angle whose vertex is on the top of the circle (on the circumference of the circle) and whose ends are chords in a circle.

B - Angle inscribed in a circle

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Inscribed angle in a circle

We are here to define for you what an inscribed angle in a circle is. Also, to give you tips to remember its definition and characteristics in the most logical way.
Before talking about the inscribed angle in the circle, let's take a moment to look at its name - inscribed angle.
Its name, lets us understand that it has a connection with the circumference and indeed it does.


Now, we can move on to the definition of an inscribed angle and it will stay in our minds thanks to logic.
What is an inscribed angle in a circle?
An inscribed angle in a circle is an angle whose vertex is on the top of the circle: on the circumference of the circle and its ends are chords in the circle.
Let's see it in the figure:

B2 - Inscribed angle in a circle

We have a circle in front of us.
We mention that an inscribed angle is an angle whose vertex is on the circle, that is, on the circumference.
and whose ends are chords in a circle.
Therefore, if you draw any two chords in a circle, they will meet at the same point on the circumference - On the circle itself, we will create an angle.
The angle that will be formed, will be an inscribed angle in the circle.
We will mark A at some point on the circle and two chords inside the circle that meet at point A A .


Now that we know what an inscribed angle in a circle is and can easily identify it,
we must know some important theorems and properties of an inscribed angle in a circle.
Shall we start?
Equal inscribed angles
When can we determine that the inscribed angles in a circle are equal?


Inscribed angles that subtend the same arc from the same side are equal to each other.

That is, if there are any chords on which inscribed angles from the same side lean, they will be equal.

Let's see this in the figure:

B3 - Inscribed angles that lean on the same chord from the same side are equal to each other

Here is a circle and a chord AB AB .
We can see that angles 1,2,3
lean on the chord AB from the same side and therefore are equal.


Example of angles leaning on the same chord but not from the same side:

B4 - Angles that lean on the same chord and are not from the same side

We can see that angles 1 and 2 do lean on the same chord but not from the same side and therefore we cannot determine that they are equal.


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Equal inscribed angles stand opposite equal chords and equal arcs.

That is, if we are given that there are equal inscribed angles, we can determine that the chords and the arcs they intercept are also equal.
Let's see this in the figure:

B5 -Equal inscribed angles are in front of equal chords and equal arcs

Before us is a circle
if we are given that:
1=2∢1=∢2
Then we can determine that:
AB=DC AB = DC
and also
AB=DC AB = DC


Facing equal arcs in a circle, we find equal inscribed angles.

That is, if we are given equal arcs in a circle, we can determine that the inscribed angles opposite them are equal.
Let's see this in the figure:

B6 -In front of equal arcs there are equal inscribed angles

Before us is a circle.
If we are given that:
AB=CD AB = CD
then
1=2∢1=∢2


Now we will study the relationship between an inscribed angle in a circle and a central angle in a circle.
Remember that a central angle in a circle is an angle whose vertex is at the center of the circle and whose ends are radii in the circle.
Like here:

7 - Central angle in a circle


Do you know what the answer is?

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

In a circle, the inscribed angle will be half of the central angle that leans on 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.
Let's see this in the figure:

8 - The inscribed angle in the circle will be half of the central angle that leans on the same arc.


The relationship between the diameter and the inscribed angle.

Remember, the diameter is the largest radius in a circle: a line that connects 2 points on the top of the circle and passes through the center of the circle.
What is the relationship between this and the circumferential angle? Excellent that you asked.
An inscribed angle that leans on a diameter equals 90° 90° degrees.
In the same way, we can say that if any inscribed angle in a circle equals 90° 90° degrees, it leans on a diameter.
Let's see this in the figure:

9 -An inscribed angle that leans on a diameter equals 90° degrees

If the diameter AB AB
then
ACB=90° ∢ACB = 90°
in the same way, if
ACB=90° ∢ACB=90°
then
AB AB is the diameter.


Check your understanding

Two inscribed angles in a circle that subtend the same chord from different sides

Do you remember we talked about the fact that inscribed angles leaning on the same chord on one side are equal?
Now, we are talking about inscribed angles leaning on the same chord but on its two different sides. The two angles together add up to 180° 180° degrees.
Let's see this in the figure:

10 - The two inscribed angles, leaning on the same chord on each side, are equal to 180° degrees

In front of us, there is a circle and a chord AB AB
The angles ACD \sphericalangle ACD and ADB \sphericalangle ADB
are inscribed angles that lean on the same chord on its two different sides and therefore their sum will be 180° 180° .



Inscribed Angle in a Circle (Examples and Exercises with Solutions)

Exercise #1

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 first that the equation of a circle with center at O(xo,yo) O(x_o,y_o) and radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2 Let's now return to the problem and the given circle equation and examine it:

x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2 We will try to give this equation a form identical to the circle equation, that is - we will ensure that on the left side of it 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:

To do this, first let's recall again the shortened multiplication formulas for binomial squared:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2 and 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 continue, for convenience and clarity of discussion - we'll separate 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 shortened multiplication 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}} \\ It can be noticed that compared to the shortened multiplication formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

{xc4ad \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d \end{cases} Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a binomial squared form,

We will need to add to these two terms 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 in question, 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 into the binomial squared form the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression further:

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 development stages so far for the expression related to x, we'll do this now 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 perform an identical process also for the expressions related to y in the resulting equation:

(Now we'll choose the addition form of the binomial squared 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 stage, we moved the free numbers to the second side and combined similar terms,

Now that we have changed the given circle equation to the form of the general circle equation mentioned earlier, we can easily extract from the given equation 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 stage, 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}

Now let's remember that the radius of the circle, by its definition as a distance of any point on the circle from the center of the circle - is positive, and therefore we must disqualify one of the options we received for the radius, for this we will use the remaining data we haven't used yet - which is the given 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 #2

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

Video Solution

Answer

Exercise #3

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

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Answer

Exercise #4

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

20

Video Solution

Answer

2

Exercise #5

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

220

Video Solution

Answer

5

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