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

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.

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.

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.

Let's first recall 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:

$x^2-6x+y^2-4y=7$

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 formulas for squared binomial expressions:

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

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

$\underline{ x^2-6x}+y^2-4y=7$

Let's continue, for convenience and clarity - 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 squared binomial formula (we'll choose the subtraction form of the squared binomial formula since the first-degree term in the expression we're dealing with $6x$ has a negative sign):

$\underline{ x^2-6x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$We can notice that compared to the squared binomial formula (from the blue box on the right in the previous calculation) we're actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

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

We'll need to add to these two terms the term$3^2$, but we don't want to change the value of the expression, so we'll also subtract this term from the expression,

In other words, we'll add and subtract the term (or expression) needed to "complete" the squared binomial 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 squared binomial form the appropriate expression (demonstrated with colors) and in the final step we'll simplify the expression further:

$x^2-2\cdot x\cdot 3\\ x^2-2\cdot x\cdot 3\underline{\underline{+3^2-3^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{3})^2-9}\\$Let's summarize the development steps so far for the x-related expression, we'll do this now within the given equation:

$x^2-6x+y^2-4y=7 \\ x^2-2\cdot x\cdot 3+y^2-4y=7 \\ (\textcolor{red}{x})^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{3}\underline{\underline{+\textcolor{green}{3}^2-3^2}}+y^2-4y=7\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{3})^2-9+y^2-4y=7\\$We'll continue and perform an identical process for the y-related terms in the resulting equation:

$(x-3)^2-9+\underline{y^2-4y}=7\\ \downarrow\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2}=7\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2\underline{\underline{+2^2-2^2}}}=7\\ \downarrow\\ (x-3)^2-9+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=7\\ \downarrow\\ (x-3)^2-9+(\textcolor{red}{y}-\textcolor{green}{2})^2-4=7\\ \boxed{(x-3)^2+(y-2)^2=20}$

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

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

$(x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{3})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{20}}$

Therefore we can conclude that the circle's center is at point:$\boxed{(x_o,y_o)\leftrightarrow(3,2)}$ and extract the circle's radius by solving a simple equation:

$R^2=20\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\sqrt{20}}$

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

Therefore, the correct answer is answer D.

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 equations and examine it:

$x^2-22x+y^2+6y=-114\leftrightarrow O\\ x^2-8ax+y^2+4ay=-11a^2\leftrightarrow M\\$We have two circle equations, from which we can extract the centers of the circles and their radii,

We will do this using "completing the square", let's start with the first circle equation (with center at point $O$):

$x^2-22x+y^2+6y=-114\\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{11}+y^2+2\cdot y\cdot \textcolor{red}{3}=-114 \\ \downarrow\\ (x-\textcolor{blue}{11})^2-\textcolor{blue}{11}^2+(y+\textcolor{red}{3})^2-\textcolor{red}{3}^2=-114\\ (x-11)^2+(y+3)^2=16\\ \rightarrow\boxed{O(11,-3),\hspace{4pt}R_O=4}$

We'll continue with the same process for the second circle (with center at point $M$):

$x^2-8ax+y^2+4ay=-11a^2\\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{4a}+y^2+2\cdot y\cdot \textcolor{red}{2a}=-11a^2\\ \downarrow\\ (x-\textcolor{blue}{4a})^2-(\textcolor{blue}{4a})^2+(y+\textcolor{red}{2a})^2-(\textcolor{red}{2a})^2=-11a^2\\ (x-4a)^2-16a^2+(y+2a)^2-4a^2=-11a^2\\ (x-4a)^2+(y+2a)^2=9a^2\\ (x-4a)^2+(y+2a)^2=(3a)^2\\ \rightarrow\boxed{M(4a,-2a),\hspace{4pt}R_M=(\textcolor{red}{+})3a}$

(Note that since it's given that: a>0 , we chose the positive sign in the radius expression we found earlier)

Now that we have expressions for the centers of the circles, we'll use the fact that the line segment between the centers$OM$ (the line connecting the centers of the two circles) is parallel to the x-axis,

Remember that for any point along a line parallel to the x-axis, the y-coordinate remains constant.

Therefore, it must be true that:

$y_O=y_M$, therefore, we'll equate the y-coordinates of the circle centers we found and solve this equation for $a$:

$\begin{cases} O(11,-3)\\ M(4a,-2a) \end{cases}\leftrightarrow y_O=y_M\\ \downarrow\\ -2a=-3\\ \downarrow\\ \boxed{a=\frac{3} {2}}$

We'll continue and calculate using the parameter value we found and using the radius expression $M$, that we found earlier, the value of this circle's radius:

$R_M=3a\leftrightarrow a=\frac{3}{2}\\ \downarrow\\ \boxed{R_M=3\cdot\frac{3}{2}=\frac{9}{2}}$

Therefore, the correct answer is answer C.

 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. 

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

$O(x_o,y_o)$

$$and its 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:

$x^2-10ax+y^2-4yb=-20ab$

We can extract from it the circle's center and its (square) radius,

We'll do this using "completing the square":

$x^2-10ax+y^2-4yb=-20ab \\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{5a}+y^2-2\cdot y\cdot \textcolor{red}{2b}=-20ab \\ \downarrow\\ (x-\textcolor{blue}{5a})^2-(\textcolor{blue}{5a})^2+(y-\textcolor{red}{2b})^2-(\textcolor{red}{2b})^2=-20ab\\ (x-5a)^2+(y-2b)^2=(5a)^2-20ab+(2b)^2\\ \rightarrow\boxed{O(5a,2b),\hspace{4pt}R_O^2=(5a)^2-20ab+(2b)^2}$

Note that for the circle's radius squared we got an algebraic expression that includes both parameters from the problem,

Note additionally that this expression can be factored into a squared binomial (using the perfect square formula):

$R_O^2=(5a)^2-20ab+(2b)^2 \\ \downarrow\\ R_O^2=(5a)^2-2\cdot5a\cdot2b+(2b)^2\\ \boxed{ R_O^2=(5a-2b)^2}$

To summarize, we found that the center of the given circle represented by the equation and its radius (squared) are:

$\rightarrow\boxed{O(5a,2b),\hspace{4pt}R_O^2=(5a-2b)^2}$

Now let's note that there is a constraint on the parameters in the problem, let's look at the radius squared expression: $R_O^2$,

and note that although this expression: $(5a-2b)^2$ must be non-negative (since it's an expression that is squared) meaning there's no contradiction with the fact that it equals the radius squared,

still- it could be zero, which would contradict the initial given- stating that the given equation represents a circle equation, (and therefore it's impossible that the radius derived from it would be zero),

meaning we must require that:

$R_O\neq0\\ \downarrow\\ 5a-2b\neq0\\ \downarrow\\ \boxed{5a\neq2b}$

Let's return now to the expression for the circle's center that we got and note that we can conclude from the constraint we found on the parameters that:

$O(x_o,y_o)\leftrightarrow O(5a,2b)\leftrightarrow 5a\neq2b\\ \downarrow\\ \boxed{x_o\neq y_o}$

meaning- it's impossible for the circle's center point to have identical x and y coordinates!

However- if we assume that the circle's center point indeed lies on the line: $y=x$then it must satisfy it, meaning it must be true that:

$\textcolor{red}{\boxed{y_o=x_o}}$

But of course as mentioned before this is a contradiction to the constraint we found earlier(!!)

(which came from requiring that the circle's radius cannot be zero)

Therefore we conclude that there are no parameters$a,\hspace{2pt}b$for which the circle's center (represented by the given equation) lies on the line: $y=x$.

Therefore- the correct answer is answer C.

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:

$x^2+2ax+y^2-2by=0$

we can extract from it the center of the circle and its (squared) radius,

we'll do this using "completing the square":

$x^2+2ax+y^2-2yb=0 \\ \downarrow\\ x^2+2\cdot x\cdot \textcolor{blue}{a}+y^2-2\cdot y\cdot \textcolor{red}{b}=0 \\ \downarrow\\ (x-\textcolor{blue}{a})^2-\textcolor{blue}{a}^2+(y-\textcolor{red}{b})^2-\textcolor{red}{b}^2=0\\ (x+a)^2+(y-b)^2=a^2+b^2\\ \rightarrow\boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2}$

we have therefore obtained the center of the circle whose equation is given in the problem and the expression for the (squared) radius in terms of the parameters in the problem,

now let's address the first given fact in the problem which states that the circle is tangent to the y-axis,

Let's recall several things:

Remember that the absolute value of the x-coordinate of the center of a circle that is tangent to the y-axis, must equal the circle's radius,

(Similarly, remember that the absolute value of the y-coordinate of the center of a circle tangent to the x-axis, must equal the circle's radius).

These facts of course stem from the fact that the tangent to a circle is perpendicular to the radius at the point of tangency (meaning - to the line segment connecting the circle's center to the point of tangency).

Let's return then to the problem, we found that the center of the circle is at point:

$O(-a,b)$

and additionally from the fact that the circle is tangent to the y-axis, as mentioned before we can conclude that:

$|x_o|=R_O$

we'll square both sides of this equation, then we'll substitute the x-coordinate of the circle's center and the squared radius and solve the equation we get:

$|x_o|=R_O\hspace{6pt}\text{/}()^2\\ \downarrow\\ x_o^2=R_O^2\leftrightarrow \boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2} \\ (-a)^2=a^2+b^2\\ a^2=a^2+b^2\\ b^2=0\\ \downarrow\\ \boxed{b=0}$

we have therefore found that the center of the circle whose equation is given in the problem is:

$O(-a,b)\stackrel{\textcolor{orange}{b=0}}{\rightarrow} O(-a,0)$

meaning - the center of the circle lies on the x-axis,

however now it's worth noting that there is another important given fact:

a<0

we'll continue and use the inequality law which states that when multiplying an inequality by a negative number - the inequality sign reverses:

a<0 \hspace{6pt}\text{/}\cdot(-1)\\ -a>0

and therefore we can conclude that the center of the circle in question must be located on the positive part of the x-axis:

O(-a,0) \leftrightarrow-a>0\\ \downarrow\\ \boxed{x_o>0}

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:

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