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Statement. For $p,q\in\mathbb R^2$ and $r>0$, there is a regulus whose two affine rulings are exactly

$$A=\{\ell_{p,a}:a\in C(q,r)\},\qquad B=\{\ell_{b,q}:b\in C(p,r)\}, \tag{1}$$

where $C(c,r)$ is the circle with center $c$ and radius $r$.

Source. Mathialagan, published 2021 PDF, p. 21, Proposition 40, introduced on p. 20 with its Figure 3. The proof below preserves the source's two circle families, makes the omitted-line limits explicit and proves completeness without importing its Lemma 39.

Existence. Choose distinct $a_1,a_2,a_3\in C(q,r)$. The lines $\ell_{p,a_i}$ are pairwise projectively disjoint by Proposition 27, and determine the quadric $R$ of Proposition 36. For $b\in C(p,r)$, the segments $(p,b)$ and $(a_i,q)$ have common positive length $r$. Their unique proper motion sends $p$ to $a_i$ and $b$ to $q$. It is a translation exactly when $a_i-p=q-b$, namely when $b=b_i=p+q-a_i$. Otherwise it is a nonidentity rotation and $\ell_{b,q}$ meets $\ell_{p,a_i}$.

If $b$ differs from the three $b_i$, the three intersection points are distinct because the original lines are disjoint. The degree-two restriction argument in Proposition 36 puts $\ell_{b,q}$ entirely on $R$. This also holds at an exceptional $b_i$: take a sequence of nonexceptional points of the circle tending to $b_i$ and substitute the parametrization of $\ell_{b,q}$ into a defining quadratic of $R$. Each coefficient of the resulting polynomial in $z$ varies continuously with $b$ and vanishes along the sequence. All coefficients vanish at $b_i$ as well.

Thus every line of $B$ lies on $R$. They are pairwise projectively disjoint, so lie in one ruling, opposite to the original triple. For any $a\in C(q,r)$, the same motion argument shows $\ell_{p,a}$ meets all lines of $B$ except the single translation case $b=p+q-a$. Three of those intersections already put $\ell_{p,a}$ on $R$, in the other ruling. This proves containment of both families (1).

No other nonhorizontal lines. Consider a nonhorizontal line $\ell_{x,y}$ in the ruling of $A$. By Corollary 37 it intersects all but at most one member $\ell_{b,q}$ of $B$. At each intersection the corresponding rotation sends $x$ to $y$ and $b$ to $q$, so

$$|x-b|=|y-q|.$$

This holds for infinitely many $b\in C(p,r)$. If $x\ne p$, two circles with distinct centers intersect in at most two points, by the elementary argument in Lemma 34. Hence $x=p$, and then $|y-q|=r$. The line is a member of $A$.

Similarly a nonhorizontal $\ell_{x,y}$ in the ruling of $B$ intersects infinitely many $\ell_{p,a}$, forcing $|x-p|=|y-a|$ for infinitely many $a\in C(q,r)$. Hence $y=q$ and $|x-p|=r$, so it belongs to $B$.

No horizontal lines. On a horizontal spatial line the angle is fixed and its centers $o$ run along an affine line. Writing its rotations as $g_o(t)=Rt+(I-R)o$, the map $o\mapsto g_o(p)$ is an invertible affine map, since $R\ne I$. Thus its images form a planar line, which meets $C(q,r)$ in at most two points. A horizontal line therefore cannot intersect infinitely many members of $A$. By Corollary 37 it cannot belong to the opposite ruling. Replacing $g_o(p)$ by $g_o^{-1}(q)=R^{-1}q+(I-R^{-1})o$ gives the same conclusion for intersections with $B$, excluding a horizontal line in the first ruling. This completes the identification of both affine rulings.

Dependencies and source qualifications. This proof uses Propositions 20, 27, 36, Corollary 37, and the circle-intersection calculation of Lemma 34. It corrects the endpoint letters in the final paragraph of the source proof. The external seven-line premise printed as Lemma 39 is not used: the two infinite families and their completeness are proved directly.

Verification scope. Verified within the independently reviewed Theorem 3 chain, retained in the final review; the complete circle classification and its applications in Lemma 26 belong to the living Theorem 3 record.

Bears on. Problem 661.