Calculating the Radius: Solving for 'a' in a Coordinate Plane Circle Problem

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.