1950 AMC 12 Problem 49

Let $M$ be the midpoint of $BC$; then $M$ moves on a circle centered at $A$

In vector form, the midpoint relation gives $\vec C=2\vec M-\vec B$ when $A$ is the origin

Put $A$ at the origin and regard $B,$ $C,$ and the midpoint $M$ of $BC$ as vectors. Since $AM=\tfrac32,$ the point $M$ moves on a circle of radius $\tfrac32$ centered at $A.$ The midpoint relation gives $$C=2M-B.$$ Thus the locus of $C$ is the image of that circle under a dilation by $2$ followed by translation by $-B.$ It is a circle of radius $3$ centered at $-B.$

Because $AB=2,$ the point $-B$ is $4$ inches from $B$ along the ray $BA.$ Hence the correct answer is D.