# Show that four points given by vectors lay on a circle

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

Show that the four points with position vectors $\vec{r_1}$, $\vec{r_2}$, $\frac{r_2}{r_1}\vec{r_1}$, $\frac{r_1}{r_2}\vec{r_2}$ , where $r_1\neq0$ and $r_2\neq0$, lie on a circle.

I tried supposing that there exists some vector $\vec{d}$ 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 $\hat{r_1}=\hat{r_2}$.

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

The cases $\vec{r_1}||\vec{r_2}$ and $|\vec{r_1}|=|\vec{r_2}|$ they are obvious.

Let $\vec{OA}=\vec{r_1},$ $\vec{OB}=\vec{r_2},$ $\vec{OB'}=\frac{r_2}{r_1}\vec{r_1}$ and $\vec{OA'}=\frac{r_1}{r_2}\vec{r_2}.$

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

• I don't really catch why BB′||AA′ ? – fazan Aug 8 '18 at 18:46
• @fazan Because $OB'=OB$ and $OA=OA'$. Draw it! – Michael Rozenberg Aug 8 '18 at 18:47

Hint:   $|\vec r_1| = r_1 = \left|\frac{r_1}{r_2} \vec r_2\,\right|$ and $|\vec r_2| = r_2 = \left|\frac{r_2}{r_1} \vec r_1\,\right|$, so the four points define an isosceles trapezoid.

Let $\vec{OA}=\vec{r_1},$ $\vec{OB}=\vec{r_2},$ $\vec{OA'}=\frac{r_2}{r_1}\vec{r_1}$ and $\vec{OB'}=\frac{r_1}{r_2}\vec{r_2}$ then

$$OA\cdot OA'=r_1\cdot r_2$$

$$OB\cdot OB'=r_1\cdot r_2$$

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

WLOG, consider $2$-D. Let $\vec{r_1}(x_1,y_1)=\vec{OA},\vec{r_2}(x_2,y_2)=\vec{OB}$. Then: $$\frac{r_2}{r_1}\vec{r_1}=\vec{OB'}\left(x_1\sqrt{\frac{x_2^2+y_2^2}{x_1^2+y_1^2}},y_1\sqrt{\frac{x_2^2+y_2^2}{x_1^2+y_1^2}}\right),\\ \left|\vec{OB'}\right|=\sqrt{x_2^2+y_2^2}=|\vec{r_2}|=\vec{OB}\\ \frac{r_1}{r_2}\vec{r_2}=\vec{OA'}\left(x_2\sqrt{\frac{x_1^2+y_1^2}{x_2^2+y_2^2}},y_2\sqrt{\frac{x_1^2+y_1^2}{x_2^2+y_2^2}}\right),\\ \left|\vec{OA'}\right|=\sqrt{x_1^2+y_1^2}=|\vec{r_1}|=|\vec{OA}|.$$ Refer to the graph:

$\hspace{6cm}$

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

If $\vec{r_1}$ and $\vec{r_2}$ 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.