Solve the Circle Equation: x² + 8x + y² - 4y = -4 with Point A(0,2)

In the given problem, we are asked to determine where a certain point is located in relation to a given circle,

To do this, we need to first find the characteristics of the given circle, namely- its center and radius,

Let's remember first that the equation of a circle with center at point

$O(x_o,y_o)$

and radius R is:

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

Additionally, let's recall that we can easily determine whether a certain point is inside/outside the circle or on it, 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 given circle equation and examine them:

$x^2+8x+y^2-4y=-4$

Let's 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 general circle equation, meaning- we'll ensure that on the left side there will be a 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, let's first recall the shortened multiplication formulas for squared binomials:

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

and we'll deal separately with the part of the equation related to x in the equation (underlined):

$\underline{x^2+8x}+y^2-4y=-4$

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 addition form of the squared binomial formula since the term with the first power we're dealing with 8x has a positive sign):

$\underline{ x^2+8x}+y^2+6y=12 \\ \underline{ x^2+8x}\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

We can notice that compared to the shortened multiplication formula (on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we'll identify that if we want to get from these two terms (underlined in the calculation) a squared binomial form,

we'll need to add to these two terms the term

$4^2$

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,

meaning- we'll add and subtract the term (or expression) needed to "complete" to a squared binomial 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 squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll further simplify the expression:

$x^2+2\cdot x\cdot 4\\ x^2+2\cdot x\cdot4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{4})^2-16}\\$

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

$x^2+8x+y^2-4y=-4\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-4y=-4 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{4})^2-16+y^2-4y=-4\\$

We'll continue and perform an identical process for the expressions related to y in the resulting equation:

(Now we'll choose the subtraction form of the squared binomial formula since the term with the first power we're dealing with 4y has a negative sign)

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

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 extract from the given equation both the center of the given circle and its radius simply:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x+\textcolor{purple}{4})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{16}}\\ \downarrow\\ C_O:(x-(-\textcolor{purple}{4}))^2+(y\stackrel{\downarrow}{- }\textcolor{orange}{2})^2=\underline{\underline{16}}\\$

In the last stage, we made sure to get the exact form of the general circle equation - meaning- 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 :

$\boxed{O(x_o,y_o)\leftrightarrow O(-4,2)}$

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

$R^2=16\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=4}$

Meaning the circle's characteristics (its center and radius) are:

$\begin{cases} O(-4,2)\\ R=4 \end{cases}$

Now in order to determine which of the options is most correct, meaning- to understand where the given point is located:

$A(0,2)$

In relation to the given circle, all we need to do is to calculate the distance between the given point and the center of the given circle (using the distance formula between two points) and check the result in relation to the circle's radius , first-

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 given point and the center of the given circle is:

$\begin{cases} O(-4,2)\\ A(0,2) \end{cases}\\ \downarrow\\ d_{OA}=\sqrt{(-4-0)^2+(2-2)^2} \\ d_{OA}=\sqrt{16+0} =\sqrt{16} \\ \boxed{d_{OA}=4}$

Meaning we got that the distance between the given point and the center of the given circle is 4,

Let's note that the distance between the given point and the center of the given circle$d_{OA}$equals exactly the circle's radius :

$d_{OA}=R=4$

Meaning- point A is located on the given circle,

(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 circle's radius from the circle's center, therefore necessarily a point located at a distance from the circle's center equal to the circle's radius - is on the circle)

And therefore the most correct answer is answer c.