2016 AMC 10B Problem 20

The radii grow from $2$ to $3$, so the scale factor is $k = \tfrac32$. A dilation with center $Z$ sends any point $P$ to
$$P' = Z + k(P - Z), \qquad\text{so}\qquad P' - P = (k - 1)(P - Z) = \tfrac12 (P - Z).$$

Find $Z$ from the known pair $A \mapsto A'$: the displacement of $A$ is $A' - A = (3, 4)$, and it must equal $\tfrac12(A - Z)$. Hence $A - Z = (6, 8)$ and $Z = (2,2) - (6,8) = (-4, -6)$.

Now the origin: its displacement is $\tfrac12(O - Z) = \tfrac12(4, 6) = (2, 3)$, so $O$ moves to $(2, 3)$, a distance of
$$\sqrt{2^2 + 3^2} = \sqrt{13}.$$

The answer is $\boxed{\textbf{(C)}\ \sqrt{13}}$.

A dilation is determined by its center and scale factor. The factor comes free from the two radii; the center comes from one known point-image pair, because the center, $P$ and $P'$ are collinear with $ZP' = k \cdot ZP$. Once $Z$ and $k$ are known, every point's motion is the fixed formula $(k-1)(P - Z)$, which is why the origin moves a definite distance even though it is not on either circle.

Displacement is a linear function of position: $P' - P = (k-1)P + c$. Plugging in $A$: $(3,4) = \tfrac12(2,2) + c$, so $c = (2,3)$. For $P = O$ the displacement is just $c = (2,3)$, of length $\sqrt{13}$. No need to find the center at all.

• Assuming the dilation is centered at the origin, so the origin does not move (choice A); the center must be found from $A$ and $A'$.
• Reporting the distance $A$ moves, $\sqrt{3^2 + 4^2} = 5$ (choice E), instead of the distance the origin moves.