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.__