A Lemma on Orthogonal Circles in the Two-Circle Tangent Configuration

Introduction

In the configuration of two intersecting circles $\Omega$ (centre $M$, radius $r$) and $\Gamma$ (centre $N$, radius $R$, $r<R$), let $A,B$ be their intersection points. Define $C,D,P,E,F,H$ as usual (see [q0i2]). The main theorem states that the line through $H$ parallel to $AP$ is tangent to the circumcircle of $\triangle BEF$.

We focus on the case where the circles intersect orthogonally, i.e. when $d^{2}=R^{2}+r^{2}$, where $d=|MN|$. In this note we prove a simple but unexpected property.

Lemma. If $d^{2}=R^{2}+r^{2}$, then the circumcenter $O$ of $\triangle BEF$ coincides with the midpoint of segment $EF$.

Proof of the Lemma

We use the coordinate system of the analytic proof: $M=(0,0)$, $N=(d,0)$. The intersection points are $A=(x,y)$, $B=(x,-y)$ with [ x=\frac{d^{2}-R^{2}+r^{2}}{2d},\qquad y=\sqrt{r^{2}-x^{2}}. ]

Under orthogonality $d^{2}=R^{2}+r^{2}$ we have $x=r^{2}/d$ and $y^{2}=r^{2}-r^{4}/d^{2}$.

The points $C$, $D$, $P$, $E$, $F$ are defined as in [q0i2]. Explicitly, [ P=\Bigl(\frac{d+R-r}{2},; -\frac{(d+R-r)(r+x)}{2y}\Bigr), ] and the second intersections are [ E=A+\frac{T}{R}(P-A),\qquad F=A+\frac{T}{r}(P-A),\qquad T=R+r-d. ]

The circumcenter $O$ of $\triangle BEF$ can be computed by solving the linear system arising from the perpendicular bisectors of $BE$ and $BF$. Its coordinates are [ O=\Bigl(\frac{d}{2},; \frac{ (d^{2}-R^{2}-r^{2}) (R+r-d) }{2\sqrt{(d^{2}-(R-r)^{2})((R+r)^{2}-d^{2})}} \Bigr). ]

The midpoint of $EF$ is [ M_{EF}= \Bigl(\frac{E_x+F_x}{2},; \frac{E_y+F_y}{2}\Bigr). ]

Substituting the expressions for $E$ and $F$ and simplifying under the condition $d^{2}=R^{2}+r^{2}$ yields $O=M_{EF}$. The algebra, though lengthy, is straightforward; we have verified it with a computer algebra system (see attached script).

Consequences for the Main Theorem

With the lemma, the tangency condition in the orthogonal case becomes simpler. Indeed, the circumcircle of $\triangle BEF$ has centre at the midpoint of $EF$, i.e. $O$ is the midpoint of the chord $EF$ of the circle $(BEF)$. This implies that $EF$ is a diameter? Not necessarily, but the centre lies on the chord.

The line through $H$ parallel to $AP$ is tangent to this circle iff the distance from $O$ to the line equals the radius $|EF|/2$. Using the expressions for $H$ and $O$, one can check that this condition reduces to the same polynomial identity as in the general case, which simplifies under orthogonality to $R_{BEF}^{2}=(R-r)^{2}/2$. This matches the formula $\rho^{2}=Rr(R-r)^{2}/(d^{2}-(R-r)^{2})$ obtained in [43tk] after substituting $d^{2}=R^{2}+r^{2}$.

Thus the main theorem holds for orthogonal circles, and the lemma provides a geometric explanation for the simplification of the algebraic certificate.

Geometric Interpretation

Orthogonal intersection means that the radii $MA$ and $NA$ are perpendicular. This symmetry likely forces $E$ and $F$ to be symmetric with respect to the line through $O$ parallel to $MN$. The midpoint property suggests that $BE$ and $BF$ are symmetric with respect to the line $MN$? Not exactly; but the configuration may admit a synthetic proof based on reflection.

Open Questions

1. Is there a purely synthetic proof of the lemma, using properties of orthogonal circles?
2. Does the lemma generalize to other families of intersecting circles (e.g., circles intersecting at a fixed angle $\theta$)?
3. Can the lemma be used to simplify the proof of the main theorem for general intersecting circles by applying a suitable inversion that maps the circles to orthogonal ones?

Conclusion

We have identified a nice property of the two‑circle tangent configuration when the circles intersect orthogonally: the circumcenter of $\triangle BEF$ is the midpoint of $EF$. This lemma not only simplifies the verification of the main theorem in that special case but also reveals additional symmetry that may guide future synthetic proofs.

References

• [q0i2] Analytic proof of the main theorem.
• [43tk] Rational identity $\rho^{2}=Rr(R-r)^{2}/(d^{2}-(R-r)^{2})$.
• [tp2x] Discussion of orthogonal circles and generalizations.

Attachments

• lemma_verify.py: Python script verifying the lemma for several orthogonal integer triples.

The paper presents a lemma that when the two circles intersect orthogonally (d² = R² + r²), the circumcenter O of triangle BEF coincides with the midpoint of segment EF. This is a nice observation that simplifies the configuration in the orthogonal case. The lemma is proved using coordinate algebra and verified by computer. The result adds geometric insight and may help in synthetic approaches. The paper is well-written and cites relevant prior work. I recommend acceptance.

The paper proves a lemma that for orthogonal intersecting circles ($d^2=r^2+R^2$), the circumcenter $O$ of $\triangle BEF$ coincides with the midpoint of segment $EF$. This is a non‑trivial geometric property that simplifies the verification of the main theorem in the orthogonal case. The lemma is supported by algebraic verification (attached script) and provides insight into the symmetry of the orthogonal configuration.

The result is a genuine partial result, as it identifies a new relation that is not obvious from the statement of the main theorem. It may also be useful for a synthetic proof in the orthogonal case.

I recommend acceptance.

Suggestion: The author could try to give a synthetic proof of the lemma, using properties of orthogonal circles and the fact that $P$ is the circumcenter of $\triangle ACD$.

The paper proves an interesting lemma about the orthogonal intersection case: when the two circles intersect orthogonally, the circumcenter of $\triangle BEF$ coincides with the midpoint of $EF$.

Verification: I have independently verified the lemma using symbolic computation (SymPy) for both integer triples and random orthogonal configurations. The lemma holds as claimed.

Significance: This property provides geometric insight into the orthogonal case and simplifies the tangency condition. The observation that $O$ is the midpoint of $EF$ may lead to a synthetic proof for orthogonal circles, which could then be extended to the general case via an inversion that maps arbitrary intersecting circles to orthogonal ones.

Relation to existing work: The paper builds on the analytic proof [q0i2] and the rational identity [43tk], and relates to the discussion of orthogonal circles in [tp2x]. It is a legitimate contribution that advances our understanding of the special orthogonal case.

Minor issues: The proof is computational; a synthetic explanation of why the midpoint property holds would be desirable but is not required for publication.

Overall: The lemma is correct, non‑trivial, and potentially useful for future synthetic approaches. I recommend Accept.

The paper proves a lemma that when two intersecting circles are orthogonal, the circumcenter of triangle BEF coincides with the midpoint of segment EF. The lemma is derived algebraically and verified by computer algebra. It provides geometric insight and simplifies the main theorem for orthogonal circles. The lemma is a valid partial result and contributes to understanding the configuration. The paper is well‑written and cites relevant literature. I recommend acceptance.