Vtyjkyuk

I'm stuck on problem 2.10 from Vector Analysis and Cartesian Tensors by Kendall:

Show that the four points with position vectors $vecr_1$, $vecr_2$, $fracr_2r_1vecr_1$, $fracr_1r_2vecr_2$ , where $r_1neq0$ and $r_2neq0$, lie on a circle.

I tried supposing that there exists some vector $vecd$ which gives the position of the circle centre and then trying to prove that the distance from each point to this centre is equal. But I just arrive at the condition that $hatr_1=hatr_2$.

Any suggestions for another strategy, am I just messing something along the way?

The cases $vecr_1||vecr_2$ and $|vecr_1|=|vecr_2|$ they are obvious.

Let $vecOA=vecr_1,$ $vecOB=vecr_2,$ $vecOB'=fracr_2r_1vecr_1$ and $vecOA'=fracr_1r_2vecr_2.$

Thus, $BB'||AA'$, $AB'=BA'$, which says that the trapezoid $AB'BA'$ is cyclic.

Hint:   $|vec r_1| = r_1 = left|fracr_1r_2 vec r_2,right|$ and $|vec r_2| = r_2 = left|fracr_2r_1 vec r_1,right|$, so the four points define an isosceles trapezoid.

Let $vecOA=vecr_1,$ $vecOB=vecr_2,$ $vecOA'=fracr_2r_1vecr_1$ and $vecOB'=fracr_1r_2vecr_2$ then

$$OAcdot OA'=r_1cdot r_2$$

$$OBcdot OB'=r_1cdot r_2$$

therefore by Circle Power the four points belong the a circle.

WLOG, consider $2$-D. Let $vecr_1(x_1,y_1)=vecOA,vecr_2(x_2,y_2)=vecOB$. Then:
$$fracr_2r_1vecr_1=vecOB'left(x_1sqrtfracx_2^2+y_2^2x_1^2+y_1^2,y_1sqrtfracx_2^2+y_2^2x_1^2+y_1^2right),\
left|vecOB'right|=sqrtx_2^2+y_2^2=|vecr_2|=vecOB\
fracr_1r_2vecr_2=vecOA'left(x_2sqrtfracx_1^2+y_1^2x_2^2+y_2^2,y_2sqrtfracx_1^2+y_1^2x_2^2+y_2^2right),\
left|vecOA'right|=sqrtx_1^2+y_1^2=|vecr_1|=|vecOA|.$$
Refer to the graph:

$hspace6cm$

Note that $AA'||BB'$, therefore $AA'BB'$ is an isosceles trapezoid, hence cyclic.

If $vecr_1$ and $vecr_2$ are collinear and directed in the same direction, the four points are still cyclic, however, if they are oppositely directed (except being equal), then the four points are cocentric.