Completing the square in a quadratic equation

First, let's recall the principles of the "completing the square" method and its general idea:

In this method, we use the formulas for the square of a binomial in order to give an expression the form of a squared binomial,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it a form of a squared binomial,

That is, we use the formulas for the square of a binomial:

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

And we bring the expression to a squared form by adding and subtracting the missing term,

In the given problem we will first refer to the given equation:

$121x^2-44x-9=0$

First, we will try to give the expression on the left side of the equation a form that resembles the form of the right side in the abbreviated multiplication formulas mentioned, we will also identify that we are interested in the subtraction form of the abbreviated multiplication formula, this is because the non-squared term in the given expression, 44x is negative, we will continue,

First, we will deal with the two terms with the highest powers in the expression requested which is on the left side of the equation,

And we will try to identify the missing term in comparison to the abbreviated multiplication formula,

To do this- first we will present these terms in a form similar to the form of the first two terms in the abbreviated multiplication formula:

$\underline{ 121x^2-44x}-9\textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \\ \hspace{4pt}\\ \\ \downarrow\\ \underline{(\textcolor{red}{11x})^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}}-9 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

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

$\begin{cases} 11x\textcolor{blue}{\leftrightarrow}c\\ 2\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

We will 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 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 form of a squared binomial,

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

Next- we will put into the squared binomial form the appropriate expression (demonstrated with colors) and in the last stage we will further simplify the expression:

$(11x)^2-2\cdot 11x\cdot 2-9\\ (11x)^2-2\cdot11x\cdot 2\underline{\underline{+2^2-2^2}}-9\\ (\textcolor{red}{11x})^2-2\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4-9\\ \downarrow\\ (\textcolor{red}{11x}-\textcolor{green}{2})^2-4-9\\ \downarrow\\ \boxed{(11x-2)^2-13}$

Hence- we obtained the completing the square form for the given expression,

Let's summarize the development stages, we will do this now within the given equation:

$121x^2-44\sqrt{2}x-9=0 \\ (11x)^2-2\cdot 11x\cdot 2-9=0\\ (\textcolor{red}{11x})^2-2\cdot \textcolor{red}{11x}\cdot \textcolor{green}{2}\underline{\underline{+\textcolor{green}{2}^2-2^2}}-9=0\\ \downarrow\\ (\textcolor{red}{11x}-\textcolor{green}{2})^2-4-9=0\\ \downarrow\\ \boxed{(11x-2)^2-13=0}$

Now, we can isolate from this expression a simpler algebraic expression,

We will do this by moving terms and extracting a square root:

$(11x-2)^2-13=0\\ (11x-2)^2=13\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{11x-2=\pm\sqrt{13}}$

(We should remember of course that extracting a square root from both sides of the equation involves considering two possibilities - with a positive sign and with a negative sign)

Let's note now that we are interested in the value of the expression:

$11x+9=\text{?}$

Which we will easily extract from the equations that we obtained,

At this stage we will emphasize two important things:

A. We obtained two equations requiring two values with opposite signs for the same expression:

$11x-2=\pm\sqrt{13}$

However it's easy to understand that these two equations cannot be held together unless the expression equals 0, which is not the case here.

B. Due to this fact, we need to separate and solve individually in order to obtain all the possibilities for the value of the requested expression,

We will continue, and refer to each equation separately, first we will try to identify the requested expression, and then isolate it, in each equation separately:

$11x-2=\pm\sqrt{13} \\ \underline{\textcolor{blue}{11x+9}}-11=\pm\sqrt{13} \\ \downarrow\\ 11x+9-11=\sqrt{13} \rightarrow\boxed{11x+9=11+\sqrt{13}} \\ 11x+9-11=-\sqrt{13}\rightarrow\boxed{11x+9=11-\sqrt{13}} \\ \downarrow\\ \boxed{11x+9=11+\sqrt{13},\hspace{4pt}11-\sqrt{13}}$

Therefore, the correct answer is answer A.

We want to calculate the value of the expression:

$x^2+\frac{9}{x^2}=\text{?}$

based on the given equation:

$x+\frac{3}{x}=5$

but without solving it for x,

For this, let's first notice that while the given deals with terms with first power only,

in the expression we want to calculate - there are terms with second power only,

therefore we understand that apparently we need to square the expression on the left side of the given equation,

We'll remember of course the shortened multiplication formula for binomial square:

$(a\pm b)^2=a^2\pm2ab+b^2$

and we'll square both sides of the given equation, later we'll emphasize something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its proportional inverse are added):

$x+\frac{3}{x}=5 \hspace{6pt}\text{/}()^2\\ (x+\frac{3}{x})^2=5^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{3}{x}}+ \frac{3^2}{x^2}=25\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{3}+ \frac{9}{x^2}=25\\$Let's now notice that the "mixed" term in the shortened multiplication formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning - it's not dependent on the variable x, since it involves multiplication between an expression with a variable and its proportional inverse,

This fact actually allows us to isolate the desired expression from the equation we get and find its value (which is not dependent on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2+2\cdot \textcolor{blue}{3}+ \frac{9}{x^2}=25\\ x^2+6+ \frac{9}{x^2}=25\\ \boxed{x^2+\frac{9}{x^2}=19}$

Therefore the correct answer is answer C.

First, let's recall the principles of the "completing the square" method and its general idea:

In this method, we use the formulas for the square of a binomial in order to give an expression the form of a squared binomial,

This method is called "completing the square" due to the fact that in this method we "complete" a missing part of a certain expression in order to obtain from it a form of a squared binomial,

That is, we use the formulas for the square of a binomial:

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

And we bring the expression to a squared form by adding and subtracting the missing term,

In the given problem we will first refer to the given equation

$2x^2+14\sqrt{2}x-15=0$

First, we will try to give the expression on the left side of the equation a form that resembles the form of the right side in the abbreviated multiplication formulas mentioned, we will also identify that we are interested in the addition form of the abbreviated multiplication formula, this is because the term that is not squared in the given expression,:

$14\sqrt{2}x$

has a positive sign,we will continue,

First, we will deal with the two terms with the highest powers in the expression requested on the left side of the equation,

And we will try to identify the missing term in comparison to the abbreviated multiplication formula,

To do this- first we will present these terms in a form similar to the form of the first two terms in the abbreviated multiplication formula:

$\underline{ 2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \hspace{4pt}\\ \\ \underline{ (\sqrt{2})^2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \downarrow\\ \underline{(\textcolor{red}{\sqrt{2}x})^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}}-15 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

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

$\begin{cases} \sqrt{2}x\textcolor{blue}{\leftrightarrow}c\\ 7\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

we will need to add to these two terms the term $7^2$

However, 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 form of a squared binomial,

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

Then- we will put into the squared binomial form the appropriate expression (demonstrated with colors) and in the last stage we will further simplify the expression:

$(\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15\\ (\sqrt{2}x)^2+2\cdot\sqrt{2}x\cdot 7\underline{\underline{+7^2-7^2}}-15\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}+\textcolor{green}{7}^2-49-15\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64}$

Thus- we obtained the completing the square form for the given expression,

Let's summarize the development stages, we will do this now within the given equation:

$2x^2+14\sqrt{2}x-15=0 \\ (\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15=0\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}\underline{\underline{+\textcolor{green}{7}^2-7^2}}-15=0\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15=0\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64=0}$

Now, we can isolate from this expression a simpler algebraic expression,

We will do this by transferring sides and extracting a square root:

$(\sqrt{2}x+7)^2-64=0\\ (\sqrt{2}x+7)^2=64\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{\sqrt{2}x+7=\pm8}$

(We will remember of course that extracting a square root from both sides of the equation involves considering two possibilities - with a positive sign and with a negative sign)

Let's note now that we are interested in the value of the expression:

$\sqrt {2}x+5=\text{?}$

Which we can easily extract from the equations we obtained,

At this stage we will emphasize two important things:

A. We obtained two equations requiring two values with opposite signs for the same expression

$\sqrt{2}x+7=\pm8$

But it's easy to understand that these two equations cannot be held together unless the expression equals 0, which is not the case here.

B. Because of this, we need to separate and solve each one independently in order to get all the possibilities for the value of the requested expression,

We will continue, and refer to each equation separately, first we will try to identify the requested expression, and then isolate it, in each equation separately:

$\sqrt{2}x+7=\pm8 \\ \underline{\textcolor{blue}{\sqrt{2}x+5}}+2=\pm8 \\ \downarrow\\ \sqrt{2}x+5+2=8 \rightarrow\boxed{\sqrt{2}x+5=6} \\ \sqrt{2}x+5+2=-8 \rightarrow\boxed{\sqrt{2}x+5=-10} \\ \downarrow\\ \boxed{\sqrt{2}x+5=6,\hspace{4pt}-10}$

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. 

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, let's remember first that the equation of a circle with center at point

$O(x_o,y_o)$

and radius R is:

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

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 through an identical process, here we'll do it in parallel for both variables:

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

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 in relation to the radii of the circles, let's first present the data of the two circles:

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

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.