<s>[INST] 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?
 [/INST] You can just observe that the origin has the same power with respect to the two pairs of points collinear with it:
$$
|\mathbf{r}_1|\cdot \left|\frac{r_2}{r_1} \mathbf{r}_1\right| 
=r_1r_2=
|\mathbf{r}_2|\cdot \left|\frac{r_1}{r_2} \mathbf{r}_2\right|.
$$
</s>