Solve Circle Equation x²-8x+y²-6y=24 Using Completing the Square Method

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

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

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

$x^2-8x+y^2-6y=24$

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 binomial square formulas:

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

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

$\underline{ x^2-8x}+y^2-6y=24$

Let's continue, for convenience and clarity of discussion - 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 binomial square formula (we'll choose the subtraction form of the binomial square formula since the first-degree term in the expression we're dealing with$8x$ has a negative sign):

$\underline{ x^2-8x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the binomial square formula (from the right side of the blue press 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 can identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term$4^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning, we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

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

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

$x^2-2\cdot x\cdot 4\\ x^2-2\cdot x\cdot 4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4})^2-16}\\$Let's summarize the development stages so far for the x-related expression, we'll do this now within the given equation:

$x^2-8x+y^2-6y=24 \\ x^2-2\cdot x\cdot 4+y^2-6y=24 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-6y=24\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4})^2-16+y^2-6y=24\\$We'll continue and perform an identical process for the y-related terms in the resulting equation:

$(x-4)^2-16+\underline{y^2-6y}=24\\ \downarrow\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3}=24\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3\underline{\underline{+3^2-3^2}}}=24\\ \downarrow\\ (x-4)^2-16+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=24\\ \downarrow\\ (x-4)^2-16+(\textcolor{red}{y}-\textcolor{green}{3})^2-9=24\\ \boxed{(x-4)^2+(y-3)^2=49}$

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

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

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4})^2+(y-\textcolor{orange}{3})^2=\underline{\underline{49}}$

Therefore we can conclude that the circle's center is at point:$\boxed{(x_o,y_o)\leftrightarrow(4,3)}$ and extract the circle's radius by solving a simple equation:

$R^2=49\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=7}$

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

Therefore, the correct answer is answer C.