4 Some constructions

1][t]0.71

Construction 4.1   Construct $y$ so that $x\, y=1$.

This can be done using the right triangle altitude theorem.

    

1][t]0.28

Construction 4.2   Construct cycles $\pi_{1}$, $\pi_{2}$ whose shadows are symmetrical.

From Proposition 2.7, cycles $\pi_{1}$, $\pi_{2}$ are reciprocal under the homographic transformation $\psi\,:\,\psi\left(z\right)=-1/\overline{z}$. When $\pi_{1}$ is a circle away from $O$, one can draw a diameter through $O$, obtaining $\alpha,\,\beta$ (Fig. 3(a)) and then use Construction 4.1 to obtain $\gamma=\psi\left(\alpha\right)$ and $\delta=\psi\left(\beta\right)$ i.e. a diameter of $\pi_{2}$.

Fig. 3: Circles having opposite apexes

[Original figure]

[Meridian figure]

Construction 4.3   Given a cycle $\pi_{1}$, construct the points $\omega_{3},\,\omega_{4}$ whose shadows are the centers on the sphere of $shadow\left(\pi_{1}\right)$.

Construct $\pi_{2}$ as in Construction 4.2. Draw a secant circle $\pi'$ and construct the radical center of $\pi',\,\pi_{1},\,\pi_{2}$ (dotted lines in Fig. 3(a)). Project orthogonally to $\left(O\,\omega_{1}\right)$ and obtain $H$. Use again Construction 4.1 since $\overline{H\omega_{3}}^{2}=\overline{H\omega_{4}}^{2}=\overline{H\alpha}\,\overline{H\beta}$.

Fig. 3(b) is drawn in the plane that contains the $\mathbb{S}$-meridian circle through $\omega_{0}$. The small circle $shadow\left(\pi_{0}\right)$ stays in a plane perpendicular to the figure, and segment $\left[S_{\alpha},\, S_{\beta}\right]$ is its diameter. Points $S_{3},\, S_{4}$ are obtained as $\mathbb{S}\cap\left(\Omega A_{1}\right)$. A special treatement is required when $shadow\left(\pi_{1}\right)$ is a great circle. In this case $\pi_{2}=\pi_{1}$while $\mathcal{A}_{1}$ is rejected at infinity. One can see directly that $\omega_{3},\,\omega_{4}=\omega_{1}/\left(1\pm r_{1}\right)$.

In any case, the "poles" cannot be used to characterize a given cycle, since they occur by pair and are not easy to distinguish (when $O$ is inside $\pi_{0}$, so is $\pi_{1}$ and $\omega_{3}$, $\pi_{1},$$\pi_{0}$ are on the same side of $\Delta$ while $\omega_{2}$ is on the other side).

Construction 4.4   Construct a circle $\pi_{1}$ centered on $\omega_{1}$ and orthogonal to cycle $\pi_{0}$.

Draw the projective line $\left(L\right)$ through the south pole $S_{\infty}$ and either $S_{1}$ or $\Omega_{1}=\left(x_{1},y_{1},0\right)$. This line intersects the projective plane $\left(P\right)$ of $shadow\left(\pi_{0}\right)$. The intersection $\left(L\right)\cap\left(P\right)$ is the required $X_{1}=apex\left(\pi_{1}\right)$.

Proof. [Discussion] When $A_{1}$ lies strictly inside $\mathbb{S}$, we have $r_{1}^{2}<0$ and therefore circle $\pi_{1}$ is imaginary. When $\omega_{1}\in\pi_{0}$ and $\pi_{0}$ is a circle, then $\pi_{1}$ is the point-circle $\left\{ \omega_{1}\right\}$. When $\pi_{0}$ is a line and $\omega_{1}\in\pi_{0}$, then $\left(L\right)\subset\left(P\right)$ and all the circles concentric at $\omega_{1}$ are solutions. When $\pi_{0}$ is a straight line and $\omega_{1}\notin\pi_{0}$, then $\left(L\right)$ intersects $\left(P\right)$ at $S_{\infty}$ and $\pi_{1}$ is the point-circle $\left\{ \infty\right\}$. The last result is counter-intuitive, but comes from the definition of concentric circles as pencil $P\left(\left\{ \omega_{1}\right\} ,\left\{ \infty\right\} \right)$ so that $\left\{ \infty\right\}$ is "centered" everywhere. $\qedsymbol$

Construction 4.5   Construct the relatives of a given point $\omega_{0}\neq O$.

Let $\left(\omega\perp\right)$ and $\left(O\perp\right)$ be respectively, the straight line drawn in the equatorial plane, through $\omega_{0}$ orthogonally to $\left(O\,\omega_{0}\right)$. Additionally, let $u$ and $v$ be respectively $u\doteq apex\left(O\,\omega\right)=\left(y,-x,0,0\right)$ and $v\doteq apex\left(O,\perp\right)=\left(x,\, y,\,0,\,0\right)$. Then :

• [ $\Omega_{0}$] is located at $\left(x_{0},y_{0},0,1\right)$, i.e. "quite at $\omega_{0}$". Can be considered as the apex of a circle centered at $\omega_{0}$ (may be imaginary, see Definition 2.3).
• [$S_{0}$] shadow of $\omega_{0}$ and apex of $\left\{ \omega_{0}\right\}$. Defined by $\mathbb{S}\cap\left(S_{\infty}\Omega_{0}\right)=\left\{ S_{\infty},S_{0}\right\}$.
• [ $\left(\Gamma\right)$] meridian (great circle) through $S_{0}$. This circle is $shadow\left(O\omega_{0}\right)$.
• [ $\left(\gamma\right)$] drawn on $\mathbb{S}$ with $\left[S_{\infty},\, S_{0}\right]$ as diameter. This circle is $shadow\left(\omega_{0}\perp\right)$.
• [$D$] intersection of tangents to $\left(\Gamma\right)$ at $S_{0}$ and $S_{\infty}$. Point $D$ is the conjugate of $\left(\gamma\right)$ respective to $\mathbb{S}$ and therefore $D=apex\left(\omega\,\perp\right)$.
• [ $\left(D\, u\right)$] apex line of pencil "all lines through $\omega_{0}$" since it contains $D=apex\left(\omega_{0}\,\perp\right)$ and $u\doteq apex\left(O\,\omega_{0}\right)$
• [ $\left(S_{0}\, S_{\infty}\right)$] apex line of pencil "all circles concentric to $\omega_{0}$" - the dual of $\left(D\, u\right)$
• [ $\left(S_{0}\, D\right)$] apex line of pencil "all cycles orthogonal to $\left(O\,\omega\right)$ at $\omega$" since it contains $apex\left(\omega\right)$ and $apex\left(\omega\,\perp\right)$
• [ $\left(S_{0}\, u\right)$] apex line of pencil "all cycles tangent to $\left(O\,\omega\right)$ at $\omega$" - the dual of $\left(S_{0}\, D\right)$
• [ $\left(S_{0}\, D\, u\right)$] generated by the last two lines, this is the apex plane of bundle "all cycles through $\omega_{0}$".
• [ $\left(S_{0}\, v\right)$] is the apex line of the pencil whose limit points are $\omega$ and $-\omega$ since, obviously, $apex\left(-\omega_{0}\right)\in\left(S_{0}\, v\right)$. The radical axis of this pencil is $\left(O\,\perp\right)$ and the apex line of pencil "all circles through $\omega_{0}$ and $-\omega_{0}$" is the line through $u$ and $\left(0,0,???\right)$

Fig. 4: Relatives of a given point