Perpendicular to a chord from the center of a circle - Examples, Exercises and Solutions

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.

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.

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.

Remember that a radius is a line segment connecting the center of a circle to any point on the circle itself.

In drawing C we can see that the line coming from the center of the circle indeed connects to a point on the circle itself, while in the other drawings the lines don't touch any point on the circle.

Therefore, C is the correct drawing.

Remember that a radius is a line segment connecting the center of the circle to a point that lies on the circle itself.

In drawing A, the line doesn't touch any point on the circle itself.

In drawing B, the line doesn't pass through the center of the circle.

We can see that in drawing C, the line that extends from the center of the circle is indeed connected to a point on the circle itself.

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

$x^2-6x+y^2-4y=7$

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that the right side contains the sum of two squared binomial expressions, one for x and one for y. We'll do this using the "completing the square" method:

For this, first let's recall the formulas for squared binomial expressions:

$(c\pm d)^2=c^2\pm2cd+d^2$

and let's deal separately with the x-related part of the equation (underlined):

$\underline{ x^2-6x}+y^2-4y=7$

Let's continue, for convenience and clarity - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the squared binomial formula (we'll choose the subtraction form of the squared binomial formula since the first-degree term in the expression we're dealing with $6x$ has a negative sign):

$\underline{ x^2-6x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the squared binomial formula (from the blue box on the right in the previous calculation) we're actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we can identify that if we want to get a squared binomial form from these two terms (underlined in the calculation),

We'll need to add to these two terms the term$3^2$, but we don't want to change the value of the expression, so we'll also subtract this term from the expression,

In other words, we'll add and subtract the term (or expression) needed to "complete" the squared binomial form,

In the next calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next, we'll put into squared binomial form the appropriate expression (demonstrated with colors) and in the final step we'll simplify the expression further:

$x^2-2\cdot x\cdot 3\\ x^2-2\cdot x\cdot 3\underline{\underline{+3^2-3^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{3})^2-9}\\$Let's summarize the development steps so far for the x-related expression, we'll do this now within the given equation:

$x^2-6x+y^2-4y=7 \\ x^2-2\cdot x\cdot 3+y^2-4y=7 \\ (\textcolor{red}{x})^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{3}\underline{\underline{+\textcolor{green}{3}^2-3^2}}+y^2-4y=7\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{3})^2-9+y^2-4y=7\\$We'll continue and perform an identical process for the y-related terms in the resulting equation:

$(x-3)^2-9+\underline{y^2-4y}=7\\ \downarrow\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2}=7\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2\underline{\underline{+2^2-2^2}}}=7\\ \downarrow\\ (x-3)^2-9+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=7\\ \downarrow\\ (x-3)^2-9+(\textcolor{red}{y}-\textcolor{green}{2})^2-4=7\\ \boxed{(x-3)^2+(y-2)^2=20}$

In the last step, we moved the free numbers to the other side and grouped similar terms,

Now that we've transformed the given circle equation into the general circle equation form mentioned earlier, we can easily extract both the center of the given circle and its radius from the given equation:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{3})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{20}}$

Therefore we can conclude that the circle's center is at point:$\boxed{(x_o,y_o)\leftrightarrow(3,2)}$ and extract the circle's radius by solving a simple equation:

$R^2=20\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\sqrt{20}}$

(remembering that the circle's radius by definition is a distance from any point on the circle to its center - is positive),

Therefore, the correct answer is answer D.

 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$(4</span><span class="katex">a)^2$, 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 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.