Let the radical axis of the (non-concentric) circles $S_A \equiv (A),\ S_B \equiv (B)$ with radii $r_A \ge r_B$ meet their center line AB at a point R and let M be the midpoint of the segment AB. Since the point R has the same power to both the circles (A), (B), $(RA + r_A)(RA - r_A) = (RB + r_B)(RB - r_B),\ \ RA^2 - r_A^2 = RB^2 - r_B^2$ Using oriented line segments, when factoring the difference of the 2 squares, $r_A^2 - r_B^2 = RA^2 - RB^2 = (AR + RB)(AR - RB) =$ $= AB\ [(AM + MR) - (MB - MR)] = 2\ AB \cdot MR$ Let P be the center of a circle (P) with radius r intersecting both the circles (A), (B) at diametrally opposite points $D_A, E_A$ and $D_B, E_B$, respectively. The center A of the circle (A) is on the radical axis of the circles (A), (P), and similarly, the center B of the circle (B) is on the radical axis of the circles (B), (P). Hence, $r^2 - PA^2 = (r - PA)(r + PA) = AD_A \cdot AE_A = r_A^2$ $r^2 - PB^2 = (r - PB)(r + PB) = BD_B \cdot BE_B = r_B^2$ $r^2 = PA^2 + r_A^2 = PB^2 + r_B^2,\ \ PB^2 - PA^2 = r_A^2 - r_B^2$ Let Q be the foot of a normal from the point P to the center line AB. From the right angle triangles $\triangle APQ, \triangle BPQ$, $PQ^2 = PA^2 - QA^2 = PB^2 - QB^2,\ \ PB^2 - PA^2 = QB^2 - QA^2$ Using oriented line segments when factoring the difference of the 2 squares, it follows that $r_A^2 - r_B^2 = QB^2 - QA^2 = (QB + AQ)(QB - AQ) =$ $= AB\ [(QM + MB) - (AM - QM)] = 2\ AB \cdot QM = -2\ AB \cdot MQ$ As a result, we see that the position of the foot Q of the normal from P to the center line AB does not depend on the point P. Hence, the locus of the centers of circles (P) intersecting the circles (A), (B) at diametrally opposite points is the normal to the center line AB at a fixed point Q. Moreover, it is clear that MQ = -MR, i.e., this normal is a reflection of the radical axis of the circles (A), (B) in the perpendicular bisector of the segment AB. If P is an arbitrary point on the locus normal $QP \perp AB$, the radius r of the circle (P) is obtained by erecting normals to the segments PA or PB at the points A, B intersecting the circles (A), (B) at diametrally opposite points $D_A, E_A$ and $D_B, E_B$ and $r = PD_A = PE_A = PD_B = PE_B$. Let the circle (P) intersect the center line AB at points U, V, so that the points U, A, B, V follow on the line AB in this order. $QU^2 = PU^2 - PQ^2 = r^2 - (PA^2 - QA^2) = r_A^2 + QA^2 = r_A^2 + (MA - QM)^2 =$ $= r_A^2 + \left(\frac{AB}{2} - \frac{r_A^2 - r_B^2}{2\ AB}\right)^2 = r_A^2 + \frac{AB^2}{4} - \frac{r_A^2 - r_B^2}{2} + \frac{r_A^4 - 2r_A^2r_B^2 + r_B^4}{4\ AB^2} =$ $= \frac{AB^2}{4} + \frac{r_A^2 + r_B^2}{2} + \frac{r_A^4 + 2r_A^2r_B^2 + r_B^4}{4\ AB^2} - \frac{r_A^2r_B^2}{AB^2} = \frac{(AB^2 + r_A^2 + r_B^2)^2 - 4 r_A^2r_B^2}{4 AB^2}$ which is an expression symmetrical with respect to $r_A, r_B$ (as expected) and completely independent on the position of the point P on the locus normal QP. Since $QP \perp AB \equiv UV$, QV = -QU. Therefore, all circles intersecting the circles (A), (B) in diametrally opposite points (i.e., centered on the locus normal) intersect the center line AB at the same 2 points U, V, i.e., they belong to the same pencil. If the (pairwise non-concentric) circle $S_A \equiv (A),\ S_B \equiv (B),\ S_C \equiv (C)$ have their pairwise radical axes intersecting at the radical center, their centers A, B, C are not collinear. The locus normals of the circle pairs (A), (B) and (B), (C) then intersect at a single point P, the center of a unique circle (P) intersecting the circles (A), (B), (C) at diametrally opposite points. By the transitivity of equivalence, the locus normal of the circle pairs (C), (A) also passes through the intersection P. If the circles (A), (B), (C) have their radical axes parallel, their centers A, B, C are collinear. Their pairwise locus normals are then also parallel, i.e., they intersect at infinity and the circle intersecting all 3 circles (A), (B), (C) at diametrally opposite points has infinite radius, i.e., it is a line, obviously their common center line $AB \equiv BC \equiv CA$, but this is not a proper circle. A proper circle can exist only if the locus normals of the circle pairs (A), (B) and (B), (C) coincide and by the transitivity of equivalence, the locus normal of the circle pair (C), (A) then coincides with the first two. In this case, we have a pencil of circles with the base points $U, V \in AB$ found earlier, intersecting all 3 circles (A), (B), (C) at diametrally opposite points.

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$r_i$ is the radius of circle $i$. 1. Collinear ==> No unique circle Note that if $A$, $B$, $C$ are collinear, then given one circle $\omega$ that cuts $S_A$, $S_B$, $S_C$ diametrically, we can always find another by reflecting $\omega$ over line $ABC$. 2. No unique circle ==> Collinear Say there are two circles $\omega_1$ and $\omega_2$. Consider the power of $A$, $B$, $C$ with respect to $\omega_1$ and $omega_2$. Because $\omega_1$ and $omega_2$ diametrically cur $S_A$, $S_B$, $S_C$, we have that each of $A$, $B$, $C$ lies on the radical axis of $\omega_1$ and $omega_2$, implying that $A,B,C$ are collinear. 3. >1 circle==> two fixed points Note that, by the above, $A,B,C$ must be collinear. Now, if a circle $\omega_1$ diametrically cuts $S_A,S_B,S_C$, then it must intersect the line $ABC$ in two points (Otherwise, it could not pass through two diametrically opposite points on any of $S_A,S_B,S_C$ as $ABC$ is their line of centers). Further note that all the circles that diametrically cut $S_A,S_B,S_C$ form a pencil. Thus, the pencil passes through two fixed points $X,Y$ on $AB$. We have by PoP on each of $A,B,C$ with respect to $\omega_1$ that $XA*AY=r_A^2$ and similarly for $B,C$. Note that, if $B$ is to the right of $A$, we get that \begin{align*} &(XB)(BY)=r_B^2\\ &\Rightarrow (XA+AB)(AY-AB)=r_B^2\\ &\Rightarrow (XA)(AY)+AB(AY-AX)-AB^2=r_B^2\\ &\Rightarrow AY-AX=\frac{r_B^2+AB^2-r_A^2}{AB} \end{align*} implying that $AY-AX$ and $(AY)(AX)$ are known values. We can thus solve explicitly for them and therefore find $X,Y$'s relation to $A,B,C$.