imc

The center of dilation must lie on the line $A{A}^{\prime }$, which can be expressed as $y=\frac{4x}{3}-\frac{2}{3}$. 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 $\frac{3}{2}$, which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\frac{3}{2}\left(2-x\right)=5-x$ (for the $x$-coordinates), and $\frac{3}{2}\left(2-y\right)=6-y$. 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 $\left(\frac{3}{2}(a+4)-4,\frac{3}{2}(b+6)-6\right)$, 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.