# Perpendicular to a chord from the center of a circle

🏆Practice the parts of a circle

The perpendicular to the chord comes out of the center of the circle, intersecting the chord, the central angle in front of the chord and the arc in front of the chord.
Moreover, if there is a section that comes out from the center of the circle and crosses the chord, it will also be perpendicular to the chord.

We are here to present the properties of the perpendicular from the center of the circle to the chord.
First, we will remember that the perpendicular is a line that forms a $90°$ degree angle.
Let's see it in the illustration:

In front of us, there is a circle.
We will mark the center of the circle with a letter $A$
Our chord will be blue and will be called $BC$.
The vertical, which comes out from the center of the circle and will be perpendicular to the chord $BC$.
We will mark it in red and call it $AD$.

## Test yourself on the parts of a circle!

Is it correct to say circumference?

Great! Now that we understand exactly what perpendicular from the center of the circle is, we'll move on to its characteristics.
You can use these characteristics without proving them.

## Characteristics of the Perpendicular to the Chord from the Center of the Circle

The perpendicular from the chord, which comes out from the center of the circle:

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### 1) Bisects the Chord

Let's see this in the illustration:

The perpendicular $AD$ cuts the chord $BD$ in half
So that:
$BD=DC$

### 2) Bisects the angle in front of the chord

Let's see this in the illustration:

The perpendicular $AD$ bisects the central angle
$∡CAB$
So that:
$∢A1=∢A2$

Do you know what the answer is?

### 3) Bisects the arc that is in front of the chord

Let's see this in the illustration:

The perpendicular $AD$ crosses the arc in front of the chord.
So that:

$BE=EC$

Similarly, these characteristics also work in reverse, and we can say that if there is a straight line that comes out from the center of the circle and crosses the chord, then it is perpendicular to the chord.

In conclusion, a perpendicular line, coming from the center of the circle, crosses the chord, the relevant central angle, and the arc opposite the chord.
We will see all the features in an illustration:

If $AD$ is perpendicular to $BC$
Then
$∢A1=∢A2$
$BD=DC$
$BE=EC$

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## Perpendicular to the chord from the center of the circle (Examples and exercises with solutions)

### Exercise #1

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.

Inside

### Exercise #2

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.

greater

### Exercise #3

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

### 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.

### Exercise #4

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.

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