Complete the Square for x²-6x+y²-4y=7: Find Circle's Center & Radius

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.