_{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 $a$.}

Question 1

_{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 \( a \).}

Question 2

_{Common Geometric Shapes}

\( C_O,\hspace{6pt}C_M \)

_{Which of them are O and M in the figure:}

\( C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\ \)

_{How many sides do they have?}

Question 3

_{Given Data: }\( A(0,2) \)

_{And the result in the table is }

\( O \)_{and its explanation:}

\( x^2+8x+y^2-4y=-4 \)

_{According to the given data:}

_{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 $a$.}

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

$(x-x_o)^2+(y-y_o)^2=R^2$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:

L**et'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,

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.__

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

_{Common Geometric Shapes}

$C_O,\hspace{6pt}C_M$

_{Which of them are O and M in the figure:}

$C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\$

_{How many sides do they have?}

In the given problem, we are asked to determine __where__ the center of a certain circle is located __in relation__ to the other circle,

To do this, we need to find **first** the characteristics of the given circles, that is - their** center coordinates and their radius**,

$O(x_o,y_o)$

** and radius** R is:

$(x-x_o)^2+(y-y_o)^2=R^2$

In addition, let's remember that we can easily determine whether a certain point is inside/outside or on a given circle by calculating the distance of the point from the center of the circle in question and comparing the result to the given circle's radius,

** Let's now return to the problem and the equations of the given circles** and examine them:

$C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\$

Let's start with **the first circle:**

$C_O: x^2-4x+y^2+6y=12$

and find its center and radius, we'll do this using the "completing the square" method,

We'll try to give this equation a **form identical to the form of the circle equation**, that is - we'll ensure that on the left side there will be the sum of two squared binomial expressions, one for x and one for y, we'll 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):

$\underline{ x^2-4x}+y^2+6y=12$

**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 with the first power 4x in the expression we're ** dealing with** is negative):

$\underline{ x^2-4x}+y^2+6y=12 \\ \underline{ x^2-4x}\textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

It can be noticed that compared to the shortened multiplication formula (on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 2\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

we'll need to add to these two terms the term

$2^2$

__However, we don't want to change the value of the expression in question, and therefore - we'll also subtract this term from the expression,__

that is - __we'll 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 insert into the binomial squared form** the appropriate expression (highlighted using colors) and in the last stage we'll simplify the expression further:

$x^2-2\cdot x\cdot 2\\ x^2-2\cdot x\cdot2\underline{\underline{+2^2-2^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{2})^2-4}\\$

**Let's summarize** the development stages so far for the expression related to x, we'll do this now **within the given circle equation**:

$C_O: x^2-4x+y^2+6y=12 \\ C_O: \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{2}\underline{\underline{+\textcolor{green}{2}^2-2^2}}+y^2+6y=12 \\ \downarrow\\ C_O:(\textcolor{red}{x}-\textcolor{green}{2})^2-4+y^2+6y=12$

We'll continue and perform an ** identical** process 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 with the first power 6y in the expression ** we're dealing with** is positive)

$(x-2)^2-4+\underline{y^2+6y}=12\\ \downarrow\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3}=12\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3\underline{\underline{+3^2-3^2}}}=12\\ \downarrow\\ (x-2)^2-4+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=12\\ \downarrow\\ ( (x-2)^2-4+(\textcolor{red}{y}+\textcolor{green}{3})^2-9=12\\ C_O:\boxed{ (x-2)^2+(y+5a)^2=25}$

In the last stage, we moved the free numbers to the other side and combined similar terms,

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

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x-\textcolor{purple}{2})^2+(y+\textcolor{orange}{3})^2=\underline{\underline{25}}\\ \downarrow\\ C_O:(x-\textcolor{purple}{2})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{3}))^2=\underline{\underline{25}}\\$

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 (highlighted by arrow)

**Therefore we can conclude that the center of the circle is at point:**

$\boxed{O(x_o,y_o)\leftrightarrow O(2,-3)}$

and extract the **circle's radius** by solving a simple equation:

$R^2=25\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=5}$

Let's summarize the information so far:

$C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases}$

Now let's approach ** the equation of the given second circle** and find its center and radius

$C_M: x^2+2x+y^2-2y=7 \\ \downarrow\\ C_M: x^2+2\cdot x\cdot 1+y^2-2\cdot y\cdot 1=7 \\\\ \downarrow\\ C_M: (x+1)^2-1^2+(y-1)^2-1^2=7 \\ C_M:\boxed{ (x+1)^2+(y-1)^2=9} \\ \downarrow\\ C_M:\boxed{ (x-(1))^2+(y-1)^2=3^2} \\\\$

Therefore we'll conclude that the circle's center and radius are:

$C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}$

Now in order to determine which of the options is the most correct, that is - **to understand where the centers of the circles are in relation to the circles themselves**, all we need to do is **calculate the distance between the centers of the circles** (** using the distance formula between two points**) and check the result

$C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases},\hspace{6pt} C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}$

Let's remember that ** the distance between two points in a plane** with coordinates:

$A(x_A,y_A),\hspace{6pt}B(x_B,y_B)$

is:

$d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$

And therefore, __the distance between the centers of the circles is:__

$d_{OM}=\sqrt{(2-(-1))^2+(-3-1)^2} \\ d_{OM}=\sqrt{9+16} =\sqrt{25} \\ \boxed{d_{OM}=5}$

__That is, we got that the distance between the centers of the circles is 5,__

Let's note that the distance between the centers of the circles

$d_{OM}$

** is equal** exactly to the radius of the circle

$C_O$

__That is - point M is on the circle__

$C_O$

__(And this follows from the definition of a circle as the set of all points in a plane that are at a distance equal to the radius of the circle from the center of the circle, therefore necessarily a point at a distance from the center of the circle equal to the radius of the circle - is on the circle)__

In addition, let's note that the distance between the centers of the circles

$d_{OM}$

** is greater than** the radius of the circle

$C_M$

__That is - point O is outside the circle__

$C_M$

__Therefore, the most correct answer is answer B.__

_{Measure angle }$C_O$_{ opposite to angle }$C_M$

_{Given Data: }$A(0,2)$

_{And the result in the table is }

$O$_{and its explanation:}

$x^2+8x+y^2-4y=-4$

_{According to the given data:}

In the given problem, we are asked to determine __where__ a certain point is located __in relation__ to a given circle,

To do this, we need to **first** find the characteristics of the given circle, namely- its** center and radius**,

** Let's remember** first that__ the equation of a circle with center at point__

$O(x_o,y_o)$

** and radius **R is:

$(x-x_o)^2+(y-y_o)^2=R^2$

Additionally, let's recall that we can easily determine whether a certain point is inside/outside the circle or on it, __by calculating the distance of the point from the center of the circle in question and comparing the result to the given circle's radius,__

** Let's now return to the problem and the given circle equation** and examine them:

$x^2+8x+y^2-4y=-4$

Let's find its center and radius, we'll do this using the "completing the square" method,

We'll try to give this equation a **form identical to the general circle equation**, meaning- we'll ensure that on the left side there will be a 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, **let's first recall** the __shortened multiplication formulas for squared binomials:__

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

and we'll deal ** separately **with the part of the equation related to x in the equation (underlined):

$\underline{x^2+8x}+y^2-4y=-4$

**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 **addition** form of the squared binomial formula since the term with the first power ** we're dealing with** 8x has a positive sign):

$\underline{ x^2+8x}+y^2+6y=12 \\ \underline{ x^2+8x}\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

We can notice that compared to the shortened multiplication formula (on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

we'll need to add to these two terms the term

$4^2$

__However, we don't want to change the value of the expression in question, and therefore- we'll also subtract this term from the expression,__

meaning-__ we'll add and subtract the term (or expression) needed to "complete" to a squared binomial form__,

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

Next- **we'll insert into the squared binomial form** the appropriate expression (highlighted with colors) and in the last stage we'll further simplify the expression:

$x^2+2\cdot x\cdot 4\\ x^2+2\cdot x\cdot4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{4})^2-16}\\$

**Let's summarize** the development stages so far for the expression related to x, we'll do this now **within the given circle equation**:

$x^2+8x+y^2-4y=-4\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-4y=-4 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{4})^2-16+y^2-4y=-4\\$

We'll continue and perform an ** identical** process for the expressions related to y in the resulting equation:

(Now we'll choose the **subtraction** form of the squared binomial formula since the term with the first power ** we're dealing with** 4y has a negative sign)

$(x+4)^2-16+\underline{y^2-4y}=-4\\ \downarrow\\ (x+4)^2-16+\underline{y^2-2\cdot y \cdot 2}=-4\\ (x+4)^2-16+\underline{y^2-2\cdot y \cdot 2\underline{\underline{+2^2-2^2}}}=-4\\ \downarrow\\ (x+4)^2-16+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=-4\\ \downarrow\\ (x+4)^2-16+(\textcolor{red}{y}-\textcolor{green}{2})^2-4=-4\\ \boxed{ (x+4)^2+(y-2)^2=16}$

In the last stage, we moved the free numbers to the other side and combined similar terms,

Now that ** we've changed the given circle equation to the form of the general circle equation** mentioned earlier, we can extract from the given equation both the center of the given circle and its radius simply:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x+\textcolor{purple}{4})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{16}}\\ \downarrow\\ C_O:(x-(-\textcolor{purple}{4}))^2+(y\stackrel{\downarrow}{- }\textcolor{orange}{2})^2=\underline{\underline{16}}\\$

In the last stage, we made sure to get the exact form of the general circle equation - meaning- where only subtraction is performed within the squared expressions (highlighted by arrow)

**Therefore we can conclude that the center of the circle is at point :**

$\boxed{O(x_o,y_o)\leftrightarrow O(-4,2)}$

And extract the **circle's radius** by solving a simple equation:

$R^2=16\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=4}$

Meaning the circle's characteristics (its center and radius) are:

$\begin{cases} O(-4,2)\\ R=4 \end{cases}$

Now in order to determine which of the options is most correct, meaning- **to understand where the given point is located:**

$A(0,2)$

**In relation to the given circle**, all we need to do is to **calculate the distance between the given point and the center of the given circle** (** using the distance formula between two points**) and check the result

Let's remember that ** the distance between two points in a plane** with coordinates :

$A(x_A,y_A),\hspace{6pt}B(x_B,y_B)$

is:

$d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$

And therefore, __the distance between the given point and the center of the given circle is:__

$\begin{cases} O(-4,2)\\ A(0,2) \end{cases}\\ \downarrow\\ d_{OA}=\sqrt{(-4-0)^2+(2-2)^2} \\ d_{OA}=\sqrt{16+0} =\sqrt{16} \\ \boxed{d_{OA}=4}$

__Meaning we got that the distance between the given point and the center of the given circle is 4,__

Let's note that the distance between the given point and the center of the given circle$d_{OA}$** equals** exactly the circle's radius :

$d_{OA}=R=4$

** Meaning- point **A

__(This follows from the definition of a circle as the set of all points in a plane that are at a distance equal to the circle's radius from the circle's center, therefore necessarily a point located at a distance from the circle's center equal to the circle's radius - is on the circle)__

__And therefore the most correct answer is answer c.__

Table A on the given result