_{Given the circle with the equation:}

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

_{and its center at point O in the second quadrant,}

_{$a\neq0$}

_{Use the completing the square method to find the center of the circle and its radius using}

_{$a$}

**Let's recall** first that __the equation of a circle with center at __$O(x_o,y_o)$__ 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:

$x^2-8ax+y^2+10ay=-5a^2$We will try to give this equation a **form identical to the circle equation**, that is - we will ensure that on the left side of it there will be the sum of two squared binomial expressions, one for x and one for y, we will 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):

$\underline{ x^2-8ax}+y^2+10ay=-5a^2$**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 in the first power __we are dealing with __$8ax$ has a negative sign):

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

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d
\end{cases}$ Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a binomial squared form,

We will need to add to these two terms the term$(4</span><span class="katex">a)^2$, __but we don't want to change the value of the expression in question, 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 into the binomial squared form** the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression further:

$x^2-2\cdot x\cdot 4a\\ x^2-2\cdot
x\cdot4a\underline{\underline{+(4a)^2-(4a)^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot
\textcolor{green}{4a}+(\textcolor{green}{4a})^2-16a^2\\
\downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2}\\$**Let's summarize** the development stages so far for the expression related to x, we'll do this now **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 perform an __identical__ process also 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 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 stage, we moved the free numbers to the second side and combined similar terms,

Now that __we have changed the given circle equation to the form of the general circle equation__ mentioned earlier, we can easily extract from the given equation 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 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 (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:

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

Now let's remember that the radius of the circle, by its definition __as a distance__ of any point on the circle from the center of the circle - is positive, and therefore __we must disqualify one of the options we received for the radius__, for this we will use the remaining data we haven't used yet - which is the given __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)

That is, 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.__

_{$O(4a,-5a),\hspace{4pt}R=-6a$}