Prob. 2.10 in Bourne & Kendall: These four points lie on a circle Here is Prob. 2.10 in the book Vector Analysis and Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:

Show that the four points with position vectors
$$
\mathbf{r}_1, \mathbf{r}_2, \frac{r_2}{r_1} \mathbf{r}_1, \frac{r_1}{r_2} \mathbf{r}_2,
$$
where $\mathbf{r}_1 \neq \mathbf{0}$ and $\mathbf{r}_2 \neq \mathbf{0}$, lie on a circle.

My Attempt:

Relative to a rectangular cartesian coordinate system $Oxyz$, let $\mathbf{r}_1 = \left( a_1, b_1, c_1 \right)$ and $\mathbf{r}_2 = \left( a_2, b_2, c_2 \right)$, where
$$
a_1^2 + b_1^2 + c_1^2 > 0 \qquad \mbox{ and } \qquad a_2^2 + b_2^2 + c_2^2 > 0,
$$
and as $\mathbf{r}_1, \mathbf{r}_2, \frac{r_2}{r_1} \mathbf{r}_1, \frac{r_1}{r_2} \mathbf{r}_2$ constitute four points on the circle, so we must also have
$$
a_1^2 + b_1^2 + c_1^2  \neq a_2^2 + b_2^2 + c_2^2.
$$
As a circle is a plane figure, so we can also assume without any loss of generality that both $\mathbf{r}_1$ and $\mathbf{r}_2$ lie in the $xy$-plane so that $c_1 = c_2 = 0$. Therefore we obtain
$$
a_1^2+ b_1^2 > 0 \qquad \mbox{ and } \qquad a_2^2 + b_2^2 > 0,
$$
and also
$$
a_1^2+ b_1^2 \neq a_2^2 + b_2^2. 
$$


Let $x^2 + y^2 + 2gx + 2fy + c = 0$, $z = 0$ be the required circle. Then substituting the coordinates of the heads of the four vectors we obtain
$$
\begin{align}
2a_1 g + 2 b_1 f + c &= - \left( a_1^2+ b_1^2 \right), \\
2a_2 g + 2 b_2 f + c &= - \left( a_2^2+ b_2^2 \right), \\
2 \sqrt{ a_2^2+ b_2^2} a_1 g + 2 \sqrt{ a_2^2+ b_2^2} b_1 f + c &= a_2^2 + b_2^2, \\
2 \sqrt{ a_1^2 + b_1^2 } a_2 g + 2 \sqrt{ a_1^2 + b_1^2} b_2 f + c &= a_1^2 + b_1^2. 
\end{align}
$$

Is what I have done so far correct? If so, then what next? How to proceed from here? Or, is there a more direct solution?
 A: You can just observe that the origin has the same power with respect to the two pairs of points collinear with it:
$$
|\mathbf{r}_1|\cdot \left|\frac{r_2}{r_1} \mathbf{r}_1\right| 
=r_1r_2=
|\mathbf{r}_2|\cdot \left|\frac{r_1}{r_2} \mathbf{r}_2\right|.
$$
A: I would do it this way . . .

Let $A,B,C,D$ be given by
$$
A=\vec{r_1},\;\;\;\;
B=\frac{r_1}{r_2}\vec{r_2},\;\;\;\;
C=\vec{r_2},\;\;\;\;
D=\frac{r_2}{r_1}\vec{r_1}
$$
interpreted as vectors or points depending on the context.

Our goal is to show that $A,B,C,D$ are concyclic.

Consider $4$ cases . . .

Case $(1)$:$\;r_1=r_2$. 

Then $D=A$ and $B=C$, so 
$$
\{A,B,C,D\}=\{A,C\}
$$
and it follows that $A,B,C,D$ are concyclic. 

Case $(2)$:$\;\vec{r_1},\vec{r_2}$ are in the same direction.

Then we can write $\vec{r_2}=k\vec{r_1}$, where $k > 0$, so we get $B=A$ and $D=C$, hence
$$
\{A,B,C,D\}=\{A,C\}
$$
and it follows that $A,B,C,D$ are concyclic.

Case $(3)$:$\;\vec{r_1},\vec{r_2}$ are in opposite directions, and $r_1\ne r_2$.

Then we can write $\vec{r_2}=-k\vec{r_1}$, where $k > 0,k\ne 1$, so we get $B=-A$ and $D=kA$, hence
$$
\{A,B,C,D\}=\{\pm A,\pm kA\}
$$
which is a set of $4$ distinct collinear points.

It follows that $A,B,C,D$ are not concyclic, since a line can intersect a circle in at most two points.

Thus for case $(3)$, the problem is wrong.

Case $(4)$:$\;\vec{r_1}\not\parallel\vec{r_2}$, and $r_1\ne r_2$.

Without loss of generality, assume $r_1 < r_2$.

Then we have
\begin{align*}
\Bigl|\overrightarrow{AD}\Bigr|
&=
\Bigl|\frac{r_2}{r_1}\vec{r_1}-\vec{r_1}\Bigl|
=
\Bigl(\frac{r_2}{r_1}-1\Bigr)|\vec{r_1}|
=
\Bigl(\frac{r_2}{r_1}-1\Bigr)r_1
=
r_2-r_1
\\[4pt]
\Bigl|\overrightarrow{BC}\Bigl|
&=
\Bigl|\vec{r_2}-\frac{r_1}{r_2}\vec{r_2}\Bigl|
=
\Bigl(1-\frac{r_1}{r_2}\Bigr)|\vec{r_2}|
=
\Bigl(1-\frac{r_1}{r_2}\Bigr)r_2
=
r_2-r_1
\\[4pt]
\end{align*}
so $\Bigl|\overrightarrow{AD}\Bigr|=\Bigl|\overrightarrow{BC}\Bigr|$.

Also we have
\begin{align*}
\overrightarrow{AB}
&=
\frac{r_1}{r_2}\vec{r_2}-\vec{r_1}
=
\frac{1}{r_2}\bigl(r_1\vec{r_2}-r_2\vec{r_1}\bigr)
\\[4pt]
\overrightarrow{DC}
&=
\vec{r_2}-\frac{r_2}{r_1}\vec{r_1}
=
\frac{1}{r_1}\bigl(r_1\vec{r_2}-r_2\vec{r_1}\bigr)
\\[4pt]
\end{align*}
so $\overrightarrow{AB}\parallel\overrightarrow{DC}$.

It follows that quadrilateral $ABCD$ is an isosceles trapezoid.

But an isosceles trapezoid is a cyclic quadrilateral since opposite angles are supplementary, hence $A,B,C,D$ are concyclic.
