Solve Circle Equation: 3x^2+30x+3y^2-90y=333 Using Complete 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:

$3x^2+30x+3y^2-90y=333$$$

Let's first notice that the form of the given equation is not completely identical to the general circle equation form we mentioned, which we know, because the coefficients of the squared terms are not 1, however they are equal to each other, and therefore this equation must represent a circle, but in order to get its characteristics we must first divide both sides of the equation by the coefficient of the squared terms, 3:

$3x^2+30x+3y^2-90y=333\hspace{6pt}\text{/}:3\\ \downarrow\\ x^2+10x+y^2-30y=111$

We have thus obtained an equation equivalent to the given equation, in the correct form - meaning - where the coefficients of the squared terms are 1, from here we can continue and get the circle's characteristics:

Let's try to give this equation a form identical to the circle equation form, meaning - we'll ensure that the right side has 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 again 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):

$\underline{x^2+10x}+y^2-30y=111$

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 addition form of the binomial square formula since the first-degree term in the expression we're dealing with $10x$$$ has a positive sign):

$\underline{ x^2+10x} \textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{5}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$One can notice that compared to the binomial square formula (on the right side of the blue press in the previous calculation) we are actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 5\textcolor{blue}{\leftrightarrow}d \end{cases}$

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$5^2$, 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 next 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 using colors) and in the final stage we'll further simplify the expression:

$x^2+2\cdot x\cdot 5\\ x^2+2\cdot x\cdot 5\underline{\underline{+5^2-5^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{5}+\textcolor{green}{5}^2-25\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{5})^2-25}\\$Let's summarize the development stages so far for the expression related to x, we'll do this now within the given equation:

$x^2+10x+y^2-30y=111 \\ x^2+2\cdot x\cdot 5+y^2-30y=111 \\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{5}\underline{\underline{+\textcolor{green}{5}^2-5^2}}+y^2-30y=111\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{5})^2-25+y^2-30y=111\\$We'll continue and perform an identical process also for the expressions related to y in the equation we got:

$(x+5)^2-25+\underline{y^2-30y}=111\\ \downarrow\\ (x+5)^2-25+\underline{y^2-2\cdot y\cdot 15}=111\\ (x+5)^2-25+\underline{y^2-2\cdot y\cdot 15\underline{\underline{+15^2-15^2}}}=111\\ \downarrow\\ (x+5)^2-25+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{15}+\textcolor{green}{15}^2-225}=111\\ \downarrow\\ (x+5)^2-25+(\textcolor{red}{y}-\textcolor{green}{15})^2-225=111\\ \boxed{ (x+5)^2+(y-15)^2=361}$

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

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

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x+\textcolor{purple}{5})^2+(y-\textcolor{orange}{15})^2=\underline{\underline{361}}\\ \downarrow\\ (x-(\textcolor{purple}{-5}))^2+(y-\textcolor{orange}{15})^2=\underline{\underline{361}}\\$Therefore we can conclude that the circle's center is at point:$\boxed{(x_o,y_o)\leftrightarrow(-5,15)}$

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

$R^2=361\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=19}$(where 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 D.