1
$\begingroup$

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. enter image description here

$\endgroup$
1
$\begingroup$

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 $1/2$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $AC$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.