Distance from a chord to the center of a circle

The distance from the chord to the center of the circle is defined as the length of the perpendicular from the center of the circle to the chord.
Theorems on the distance from the center of the circle:

Chords that are equal to each other are equidistant from the center of the circle.
If in a circle, the distance of a chord from the center of the circle is less than the distance of another chord from the center of the circle, we can determine that the chord with the lesser distance is longer than the other chord.

A1 - The distance from the chord to the center of the circle

All theorems can also exist in reverse.

Detailed explanation

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Distance from the chord to the center of the circle

We already know what a chord in a circle is and of course we know what the center of the circle means.
But, have you thought about the distance between the center of the circle and the chord?
That's exactly why we're here! We will introduce you to the theorems about the distance of the chord from the center of the circle that you can use without needing to prove them.
The theorems about the distance of the chord from the center of the circle make sense, there's no reason not to understand them and remember them naturally.

First of all, it's important to know:

The distance of the chord from the center of the circle is defined for us as the length of the perpendicular line that goes from the center of the circle to the chord.
That is:

The distance is a vertical line that goes from the chord to the center of the circle

Shall we start?

The chords that are equal to each other are equidistant from the center of the circle.

Let's see this in the illustration:

We will mark $A$ as the center of the circle

If
$BC=DE$
Then
$AF=AG$

This theorem also works in reverse and therefore
if the distance of the chords from the center of the circle is equal, the chords are equal to each other.
Therefore:
If
$AF=AG$
Then
$BC=DE$

If in a circle, the distance of a chord from the center of the circle is less than the distance of another chord from the center of the circle, we can determine that the chord with the shorter distance is longer than the other chord.

This theorem can be a bit confusing.
To remember:
shorter distance - longer chord.
longer distance - shorter chord.

The truth is that you really don't have to memorize this sentence because you will see it beautifully in the illustration.
Let's see:

A3 - the chord with the shorter distance is longer than the other chord

In front of us is a circle.
We will mark - $A$ as the center of the circle.
We can clearly see that the chord $BC$ is shorter than the chord $ED$ .
We can also observe that for the shorter chord $BC$ , the distance is greater from the center of the circle
and rather to the longer chord $ED$ , shorter distance than the distance from the circle.
According to this theorem we can say that:
if
$AF<AG$
Then
$BD<ED$
This theorem also works in reverse, so we can also say that:
if
$BD<ED$
Then
$AF<AG$

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Distance from the chord to the center of the circle (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 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 take 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, that is - we will ensure that on the left side 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:

Let's recall the shortcut 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 $8ax$ has a negative sign):

$\underline{ x^2-8ax} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$ Notice that compared to the shortcut 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 isin 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)

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: