A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$. What distance does the origin $O(0,0)$, move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
The center of dilation must lie on the line $A A'$, which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$. Note that the center of dilation must have an $x$-coordinate less than $2$; if the $x$-coordinate were otherwise, then the circle under the transformation would not have an increased $x$-coordinate in the coordinate plane. Also, the ratio of dilation must be equal to $\dfrac{3}{2}$, which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\dfrac{3}{2} \left( 2 - x \right) = 5 - x$ (for the $x$-coordinates), and $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$. We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$-coordinate, and hence a lower $y$-coordinate, from our reasoning above. Solving the two equations, we get $x = -4$ and $y = - 6$. This means that any point $(a,b)$ on the plane will dilate to the point $\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)$, which means that the point $(0,0)$ dilates to $\left( 6 - 4, 9 - 6 \right) = (2,3)$. Thus, the origin moves $\sqrt{2^2 + 3^2} = \boxed{\sqrt{13}}$ units.