Solve Circle Equation: (2/7)x² + 4x + (2/7)y² + (8/7)y = 22/7

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:

$\frac{2}{7}x^2+4x+ \frac{2}{7}y^2+ \frac{8}{7}y= \frac{22}{7}$

First, let's note that the form of the given equation is not exactly identical to the general circle equation form 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,

$\frac{2}{7}$

, however since we're dealing with a fraction - we prefer to multiply instead of divide, meaning - we'll multiply both sides of the equation by

$\frac{7}{2}$

Next we'll reduce the fraction multiplications in the expression:

$\frac{2}{7}x^2+4x+ \frac{2}{7}y^2+ \frac{8}{7}y= \frac{22}{7} \hspace{6pt}\text{/}\cdot \frac{7}{2} \\ \downarrow\\ \frac{7}{2}\cdot \frac{2}{7}x^2+\frac{7}{2}\cdot4x+ \frac{7}{2}\cdot\frac{2}{7}y^2+ \frac{7}{2}\cdot\frac{8}{7}y= \frac{7}{2}\cdot\frac{22}{7} \\ \downarrow\\ x^2+14x+y^2+4y=11$

We have thus obtained an equation equivalent to the given one, 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:

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that on its right 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:

For this, first let's recall again the perfect square trinomial formulas:

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

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

$x^2+14x+y^2+4y=11 \\ \underline{ x^2+14x}+y^2+4y=11$

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 perfect square formula (we'll choose the addition form of the perfect square trinomial since the first-degree term in the expression we're dealing with $14x$ has a positive sign):

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

One can notice that compared to the perfect square formula (from the blue box on the right in the previous calculation) we are actually making the analogy:

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

Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a perfect square trinomial form,

we'll need to add to these two terms the term$7^2$, 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 perfect square trinomial,

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

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

Let's summarize the development stages so far for the expression related to x, we'll do this now within the given equation:

$x^2+14x+y^2+4y=11 \\ x^2+2\cdot x\cdot 7+y^2+4y=11 \\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{7}\underline{\underline{+\textcolor{green}{7}^2-7^2}}+y^2+4y=11\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{7})^2-49+y^2+4y=11\\$

We'll continue and perform an identical process also for the terms related to y in the equation we obtained:

$(x+7)^2-49+\underline{y^2+4y}=11\\ \downarrow\\ (x+7)^2-49+\underline{y^2+2\cdot y\cdot 2}=11\\ (x+7)^2-49+\underline{y^2+2\cdot y\cdot 2\underline{\underline{+2^2-2^2}}}=11\\ \downarrow\\ (x+7)^2-49+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=11\\ \downarrow\\ (x+7)^2-49+(\textcolor{red}{y}+\textcolor{green}{2})^2-4=11\\ \boxed{ (x+7)^2+(y+2)^2=64}$

In the last stage we moved the free numbers to the other side and grouped 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 given circle's center and its radius from the given equation:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x+\textcolor{purple}{7})^2+(y+\textcolor{orange}{2})^2=\underline{\underline{64}}\\ \downarrow\\ \big(x-(\textcolor{purple}{-7})\big)^2+\big(y-(\textcolor{orange}{-2})\big)^2=\underline{\underline{64}}\\$

Therefore we can conclude that the circle's center point is:

$\boxed{(x_o,y_o)\leftrightarrow(-7,-2)}$

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

$R^2=64\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=8}$

(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.