# Finding the locus of points which are the midpoints of segments connecting a given circle and a given point

I have been solving a problem of finding the locus of points which are midpoints of segments connecting a given circle and a given point. I found out that it is a circle with the centre in the midpoint of $AC$ which has the radius $\frac{R}{2}$, where R is the radius of the given circle. But how do I prove that? I have tried the coordinate method, and it did not give me the equation. ## 3 Answers

WLOG (i.e. for other circles/points, imitate this) your original circle is unit and the point is $(2,0)$. The points are the midpoints of the segments joining $(x,\pm\sqrt{1-x^2})$ to $(2,0)$, so they are $(X,Y)$ with $$X = \frac{x+2}{2}, Y = \frac{\pm \sqrt{1-x^2}}{2}$$

$$(X-1)^2 + Y^2 = \frac1{4}$$

So (since $Y$ is allowed both negative and positive values) this describes the circle of center $(1,0)$ and radius $1/2$.

You can do it by similar triangles. Let $P$ be the midpoint of $AC$. Then $CAD$ and $CPE$ are similar triangles, so $PE$ must be half the length of $AD$. As $D$ moves around the circle (and $E$ consequently moves), the length $AD$ doesn't change, so the length $PE$ must remain constant as well; thus $E$ traces a circle.

This transformation which takes $D$ to $E$ is just homothety with center at $C$ and $k={1\over 2}$. Since $D$ describes circle so does $E$, but with half of the radius of the first circle. The center of this new circle is midpoint of $AC$.