# Tangent to a circle

🏆Practice the parts of a circle

A tangent to a circle is a line that touches the circle at one point.

Tangent Theorem:

1) The tangent to the circle is perpendicular to the radius at the starting point

2) Every line perpendicular to the radius at its end is tangent to the circle

3) The angle between the tangent and any chord is equal to the circumferential angle that rests on that chord on the other side.

4) Two tangents to the circle that come out from the same point are equal to each other.

5) A segment that passes between the center of the circle and the point from which two tangents to the circle come out, cuts the angle between the tangents.

6) If from any point outside the circle, a tangent comes out and cuts the circle, then the product of the entire tangent on its outside is equal to the square of the tangent.

7) In the triangle that encloses the circle, the three bisectors of the angles of the triangle meet at a point in the center of the circle.

8) We can determine that a convex quadrilateral encloses a circle only if - the sum of two opposite sides in the square will be equal to the sum of the other two sides in the square.

## Test yourself on the parts of a circle!

Is it correct to say circumference?

## Tangent of a circle

Before we start learning about the many properties of a tangent to a circle, we ask ourselves:
What is a tangent of a circle?
There's nothing to fear from the word tangent. A tangent is simply a line that touches something - a tangent.
A tangent to a circle is a line that touches the circle at one point.
Let's look at the tangent in the figure:

We will notice that the tangent touches only one point on the circle and therefore does not pass through it but outside of it.
The point where the tangent touches the circle is called the point of tangency.

## Tangent Theorem

Wonderful! Now we will go to the theorems of the tangent to a circle that we can use without proving them.
To more easily remember all the tangent theorems, we will divide the tangent theorems into $4$ groups:
tangent and radius, tangent and angles, two tangents, tangent with a triangle and a quadrilateral.

We know it might be a bit scary, but don't worry, you will gradually understand the topic and see that the devil is not so terrible.
Shall we begin?

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### First group: tangent and radius

In this group, there are $2$ theorems that are essentially the same connected with the relationship between the tangent and the radius. (inverse theorem)

1) The tangent to the circle, perpendicular to the radius at the point of tangency.

That is, the tangent to the circle, the one we drew before, which touches the circle at only one point, is perpendicular to the radius of the circle, forms a right angle of $90^o$ degrees with it, at the point of tangency.
Let's see this in the figure:

If $BC$ is tangent to the circle
then
$∢ABC=90$
We can see that the tangent to the circle forms a right angle of $90^o$ degrees with the radius. This angle is formed at the point of tangency where the tangent touches the circle.

Magnificent! Now we will move on to the second theorem, the opposite of the first theorem.

2) Every line perpendicular to the radius at its endpoint is tangent to the circle.

That is, if there is a straight line that forms a $90^o$ degree angle with the radius at its endpoint, we can determine that it is tangent to the circle: it touches the circle at one point.

This is a theorem that actually proves that the straight line is tangent to the circle.
Observing the figure, we can determine that:

if

$∢ABC=90$

then

$BC$ is a tangent to the circle

Wonderful! Now we will move on to the second group.

### Second group: tangent and angles

In this group, only the Ed theorem describes the angle between a tangent and a chord.
The angle between the tangent and any chord is equal to the inscribed angle that rests on that chord from its other side.

Let's explain this theorem.

First, let's understand what is an angle between a tangent and a chord:

$AC$ is the tangent.
$BD$ is the chord.
What is the angle formed between the tangent and the chord?
$∢ADB$
We will mark it in red.

Note that another angle has been created between the tangent and the chord $∢CDB$
We will see it next.
What is the inscribed angle that rests on the same chord from the other side?
We learned that an inscribed angle is an angle whose vertex is on the top of the circle and whose legs are the chords. Let's see the peripheral angle supported on the same chord from the other side of the figure:

The angle that rests on the same chord on the other side is $∢DEB$
We will mark it in orange.
the angle between the tangent and any chord is equal to the peripheral angle that rests on that chord from the other side.
Therefore
$∢DEB =∢ADB$
Remember we said the tangent creates another angle with the chord?
Now we will refer to that:
$∢CDB$
and mark it in red.

Can you say which angle is equal to the one that rests on the same chord on the other side?
Of course!
$∢DGB$
and therefore
$∢CDB=∢DGB$

Pay attention! Sometimes in these exercises, $2$ angles between a tangent and a chord will be combined, and you will have to identify which is equal to which angle according to this theorem.
Therefore, practice it well. This is exactly which inscribed angle, the angle is equal to the angle between a tangent and a chord and you'll be ready for any scenario.
And now... we will move on to the third group.

Do you know what the answer is?

### Third group: two tangents

In this group, there are three theorems that describe the works and properties of two tangents to a circle.
1) Two tangents to the circle that start from the same point are equal to each other.

That is, if we have a circle in front of us, even if there are $2$ tangents that come out from the same point (regardless of the starting point) they will be equal to each other.
Let's see this in the figure:

We have a circle in front of us with two tangents
$AB$ and $CB$
both tangents, come out from the same point $B$.
According to the theorem when two tangents to a circle come out from the same point they are equal to each other and therefore:
$AB=CB$

2) A segment that passes between the center of the circle and the point from which two tangents to the circle emerge, cuts the angle between the tangents.

That is, if there are two tangents to the circle that come from the same point and there is a segment that connects their starting point with the center of the circle, this segment also crosses the angle between the tangents.
Observe this in the figure:

We have before us a circumference and two tangents to it that come from the same point $AB$ and $CB$
The segment $EB$ is the segment that joins the center of the circle E with the point from which the two tangents $B$ emerge.
According to the theorem, this segment is also the bisector of the angle between the tangents and therefore:
$∢CBE=∢ABE$

3) If a tangent comes out from a point outside the circle and cuts the circle, then the product of the entire cut on its exterior is equal to the square of the tangent.

Don't worry, you'll see what we mean by this theorem in the illustration and it will become much clearer:

Before us is a circle.
$AB$ is a tangent to the circle
and $BD$ is the intersection.
Both come out from the same point $B$.
Note that:
$CB$ is the exterior part of the intersection.
According to the theorem, we can determine that:

$DB*CB=AB^2$

To understand the theorem, we will give numbers to the different lengths.
Numerical example:
$AB = X$ Unknown
$CB = 1$
$DC = 3$

Can we find $AB$ the length of the tangent?
Of course:
first calculate what is the length of the entire section:
$1+3=4$
We replace in the formula and obtain that:
$4*2=X^2$
$8=X^2$
$X=2.282$
Therefore, the length of the tangent is $2.282$

Magnificent. Now we will move on to the fourth and final group.

### Fourth group: tangent with a triangle and a quadrilateral

In this group, there are two theorems and they describe the tangents to the circle when they are part of a triangle or quadrilateral that encloses the circle.

1) In a triangle that encloses the circle, the three angle bisectors of the triangle meet at a point in the center of the circle.

This theorem is simple and easy. It just describes the fact that when a circle is enclosed in a triangle, the three angle bisectors of the triangle meet at the center of the circle.
We will notice that the sides of the triangle are actually three tangents to the circle, hence the connection to the tangent.
Let's see this in the figure and understand it better:

Before us is a circle whose center is $A$.
We can notice that the circle is enclosed within a triangle.
We can also see that the three angle bisectors in the triangle meet at the center of the enclosed circle.

2) We can determine that a convex quadrilateral blocks a circle only if - the sum of two opposite sides in the quadrilateral is equal to the sum of the other two sides in the quadrilateral.

Basically, this is a condition for checking that the square blocks the circle.
We will notice that the square creates four tangents to the circle, and therefore, the connection with the tangent.
Let's see this in the figure to understand it better:

Before us, there is a circle and a convex quadrilateral.
We can determine that:

if $AB+CD=AD+BC$

then $ABCD$ Quadrilateral that blocks the circle

Excellent! Now you know all the theorems of tangents in depth and can use them from now on!

## How is the tangent of a circle calculated? (Examples and exercises with solutions)

### Exercise #1

Given the circle with the equation:

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

and its center at point O in the second quadrant,

$a\neq0$

Use the completing the square method to find the center of the circle and its radius using

$a$

### Step-by-Step Solution

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

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

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

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

$\underline{ x^2-8ax}+y^2+10ay=-5a^2$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 $8ax$ has a negative sign):

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

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d \end{cases}$ 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$(4a)^2$, 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:

$x^2-2\cdot x\cdot 4a\\ x^2-2\cdot x\cdot4a\underline{\underline{+(4a)^2-(4a)^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}+(\textcolor{green}{4a})^2-16a^2\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2}\\$Let's summarize the development stages so far for the expression related to x, we'll do this now within the given equation:

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

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4a})^2+(y+\textcolor{orange}{5a})^2=\underline{\underline{36a^2}}\\ \downarrow\\ (x-\textcolor{purple}{4a})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{5a}))^2=\underline{\underline{36a^2}}\\$ 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:$\boxed{O(x_o,y_o)\leftrightarrow O(4a,-5a)}$ and extract the radius of the circle by solving a simple equation:

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

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:

$\rightarrow \boxed{R=-6a}$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

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

2

### Exercise #5

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

5