Solving Quadratic Equations by Completing the Square: Combination with analytic geometry

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

$(x-x_o)^2+(y-y_o)^2=R^2$Now, let's now have a look at the equation for the given circle:

$x^2-8ax+y^2+10ay=-5a^2$
We will try rearrange this equation to match the circle equation, or in other words we will ensure that on the left side is the sum of two squared binomial expressions, one for x and one for y.

We will do this using the "completing the square" method:

Let's recall the short formula for squaring a binomial:

$(c\pm d)^2=c^2\pm2cd+d^2$We'll deal separately with the part of the equation related to x in the equation (underlined):

$\underline{ x^2-8ax}+y^2+10ay=-5a^2$

We'll isolate 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 shortcut formula (we'll choose the subtraction form of the binomial squared formula since the term in the first power we are dealing with is$8ax$$$, which has a negative sign):

$\underline{ x^2-8ax} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$Notice that compared to the short formula (which is on the right side of the blue arrow in the previous calculation), we are actually making the comparison:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d \end{cases}$ Therefore, if we want to get a squared binomial form from these two terms (underlined in the calculation), we will need to add the term$(4</span><span class="katex">a)^2$, but we don't want to change the value of the expression, and therefore we will also subtract this term from the expression.

That is, we will add and subtract the term (or expression) we need to "complete" to the binomial squared form,

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

Next, we'll put the expression in the squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression:

$x^2-2\cdot x\cdot 4a\\ x^2-2\cdot x\cdot4a\underline{\underline{+(4a)^2-(4a)^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}+(\textcolor{green}{4a})^2-16a^2\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2}\\$Let's summarize the steps we've taken so far for the expression with x.

We'll do this within the given equation:

$x^2-8ax+y^2+10ay=-5a^2 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4a}\underline{\underline{+\textcolor{green}{(4a)}^2-(4a)^2}}+y^2+10ay=-5a^2\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2+y^2+10ay=-5a^2\\$We'll continue and do the same thing for the expressions with y in the resulting equation:

(Now we'll choose the addition form of the squared binomial formula since the term in the first power we are dealing with $10ay$$$ has a positive sign)

$(x-4a)^2-16a^2+\underline{y^2+10ay}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a}=-5a^2\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a\underline{\underline{+(5a)^2-(5a)^2}}}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{5a}+\textcolor{green}{(5a)}^2-25a^2}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+(\textcolor{red}{y}+\textcolor{green}{5a})^2-25a^2=-5a^2\\ \boxed{(x-4a)^2+(y+5a)^2=36a^2}$In the last step, we move the free numbers to the second side and combine like terms.

Now that the given circle equation is in the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4a})^2+(y+\textcolor{orange}{5a})^2=\underline{\underline{36a^2}}\\ \downarrow\\ (x-\textcolor{purple}{4a})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{5a}))^2=\underline{\underline{36a^2}}\\$

In the last step, we made sure to get the exact form of the general circle equation—that is, where only subtraction is performed within the squared expressions (emphasized with an arrow)

Therefore, we can conclude that the center of the circle is at:$\boxed{O(x_o,y_o)\leftrightarrow O(4a,-5a)}$ and extract the radius of the circle by solving a simple equation:

$R^2=36a^2\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\pm6a}$

Remember that the radius of the circle, by its definition is the distance between any point on the diameter and the center of the circle. Since it is positive, we must disqualify one of the options we got for the radius.

To do this, we will use the remaining information we haven't used yet—which is that the center of the given circle O is in the second quadrant.

That is:

O(x_o,y_o)\leftrightarrow x_o<0,\hspace{4pt}y_o>0 (Or in words: the x-value of the circle's center is negative and the y-value of the circle's center is positive)

Therefore, it must be true that:

\begin{cases} x_o<0\rightarrow (x_o=4a)\rightarrow 4a<0\rightarrow\boxed{a<0}\\ y_o>0\rightarrow (y_o=-5a)\rightarrow -5a>0\rightarrow\boxed{a<0} \end{cases}

We concluded that a<0 and since the radius of the circle is positive we conclude that necessarily:

$\rightarrow \boxed{R=-6a}$Let's summarize:

$\boxed{O(4a,-5a), \hspace{4pt}R=-6a}$Therefore, the correct answer is answer d. 

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

To do this, we need to find first the characteristics of the given circles, that is - their center coordinates and their radius, let's remember first that the equation of a circle with center at point

$O(x_o,y_o)$

and radius R is:

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

In addition, let's remember that we can easily determine whether a certain point is inside/outside or on a given circle 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 equations of the given circles and examine them:

$C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\$

Let's start with the first circle:

$C_O: x^2-4x+y^2+6y=12$

and 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 form of the circle equation, that is - we'll ensure that on the left 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:

To do this, first let's recall again the shortened multiplication formulas for binomial squared:

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

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

$\underline{ x^2-4x}+y^2+6y=12$

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 subtraction form of the binomial squared formula since the term with the first power 4x in the expression we're dealing with is negative):

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

It can be noticed 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:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 2\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

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

$2^2$

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,

that is - we'll add and subtract the term (or expression) we need to "complete" to the binomial squared 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 binomial squared form the appropriate expression (highlighted using colors) and in the last stage we'll simplify the expression further:

$x^2-2\cdot x\cdot 2\\ x^2-2\cdot x\cdot2\underline{\underline{+2^2-2^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{2})^2-4}\\$

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

$C_O: x^2-4x+y^2+6y=12 \\ C_O: \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{2}\underline{\underline{+\textcolor{green}{2}^2-2^2}}+y^2+6y=12 \\ \downarrow\\ C_O:(\textcolor{red}{x}-\textcolor{green}{2})^2-4+y^2+6y=12$

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

(Now we'll choose the addition form of the binomial squared formula since the term with the first power 6y in the expression we're dealing with is positive)

$(x-2)^2-4+\underline{y^2+6y}=12\\ \downarrow\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3}=12\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3\underline{\underline{+3^2-3^2}}}=12\\ \downarrow\\ (x-2)^2-4+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=12\\ \downarrow\\ ( (x-2)^2-4+(\textcolor{red}{y}+\textcolor{green}{3})^2-9=12\\ C_O:\boxed{ (x-2)^2+(y+5a)^2=25}$

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 easily extract both the center of the given circle and its radius:

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x-\textcolor{purple}{2})^2+(y+\textcolor{orange}{3})^2=\underline{\underline{25}}\\ \downarrow\\ C_O:(x-\textcolor{purple}{2})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{3}))^2=\underline{\underline{25}}\\$

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

$\boxed{O(x_o,y_o)\leftrightarrow O(2,-3)}$

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

$R^2=25\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=5}$

Let's summarize the information so far:

$C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases}$

Now let's approach the equation of the given second circle and find its center and radius through an identical process, here we'll do it in parallel for both variables:

$C_M: x^2+2x+y^2-2y=7 \\ \downarrow\\ C_M: x^2+2\cdot x\cdot 1+y^2-2\cdot y\cdot 1=7 \\\\ \downarrow\\ C_M: (x+1)^2-1^2+(y-1)^2-1^2=7 \\ C_M:\boxed{ (x+1)^2+(y-1)^2=9} \\ \downarrow\\ C_M:\boxed{ (x-(1))^2+(y-1)^2=3^2} \\\\$

Therefore we'll conclude that the circle's center and radius are:

$C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}$

Now in order to determine which of the options is the most correct, that is - to understand where the centers of the circles are in relation to the circles themselves, all we need to do is calculate the distance between the centers of the circles (using the distance formula between two points) and check the result in relation to the radii of the circles, let's first present the data of the two circles:

$C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases},\hspace{6pt} C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}$

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 centers of the circles is:

$d_{OM}=\sqrt{(2-(-1))^2+(-3-1)^2} \\ d_{OM}=\sqrt{9+16} =\sqrt{25} \\ \boxed{d_{OM}=5}$

That is, we got that the distance between the centers of the circles is 5,

Let's note that the distance between the centers of the circles

$d_{OM}$

is equal exactly to the radius of the circle

$C_O$

That is - point M is on the circle

$C_O$

(And 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 radius of the circle from the center of the circle, therefore necessarily a point at a distance from the center of the circle equal to the radius of the circle - is on the circle)

In addition, let's note that the distance between the centers of the circles

$d_{OM}$

is greater than the radius of the circle

$C_M$

That is - point O is outside the circle

$C_M$

Therefore, the most correct answer is answer B.

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.