A Simple Rational Identity Underlying a Two-Circle Tangent Theorem

Introduction

In the configuration of two intersecting circles $\Omega$ (centre $M$, radius $r$) and $\Gamma$ (centre $N$, radius $R$, $r<R$) intersecting at $A$ and $B$, let $C,D,P,E,F,H$ be defined as usual. Denote by $O$ the circumcenter of $\triangle BEF$ and by $R_{BEF}$ its radius. The theorem proved in [{q0i2}] states that the line $\ell$ through $H$ parallel to $AP$ is tangent to the circle $(BEF)$; i.e. [ \operatorname{dist}(O,\ell)=R_{BEF}. ]

After carrying out the coordinate computations of [{q0i2}], both sides of this equality simplify to the same rational expression in the parameters $d,r,R$, where $d=|MN|$. This note records that expression and verifies the identity directly, without invoking the geometric construction.

The identity

Let $d>0$, $0<r<R$, and assume $|R-r|<d<R+r$ (so that the circles intersect in two distinct points). Define [ \boxed{; \rho^2(d,r,R)=\frac{R,r,(R-r)^{2}}{d^{2}-(R-r)^{2}}; }. ]

Theorem. In the configuration described above, [ R_{BEF}^{2}= \rho^2(d,r,R),\qquad \operatorname{dist}(O,\ell)^{2}= \rho^2(d,r,R). ]

Thus the tangency condition is equivalent to the equality of two quantities that both coincide with $\rho^2$.

Direct verification

The analytic proof in [{q0i2}] proceeds by computing the coordinates of all points and simplifying the difference $R_{BEF}^{2}-\operatorname{dist}(O,\ell)^{2}$ to zero. One can, however, check the identity in a purely algebraic fashion.

Let [ x_0=\frac{d^{2}-(R^{2}-r^{2})}{2d},\qquad y_0=\sqrt{r^{2}-x_0^{2}}. ]

Define the points [ \begin{aligned} A&=(x_0,y_0), & B&=(x_0,-y_0),\ C&=(-r,0), & D&=(d+R,0),\ P&=\Bigl(\frac{d+R-r}{2},;-\frac{(R+d-r)(d+r-R)(R+d+r)}{2\sqrt{\Delta}}\Bigr),\ \Delta&=(d^{2}-(R-r)^{2})((R+r)^{2}-d^{2}). \end{aligned} ]

Let $T=R+r-d$. Then [ E=A+\frac{T}{R}(P-A),\qquad F=A+\frac{T}{r}(P-A). ]

The circumcenter $O$ of $\triangle BEF$ is [ O=\Bigl(\frac{d}{2},;-\frac{d,T,(R+d+r)}{2\sqrt{\Delta}}\Bigr). ]

The orthocenter $H$ of $\triangle PMN$ is [ H=\Bigl(\frac{d+R-r}{2},;-\frac{(R+d-r)(d+r-R)T}{2\sqrt{\Delta}}\Bigr). ]

The direction vector of $AP$ is $v=P-A$. The squared distance from $O$ to the line through $H$ parallel to $AP$ is [ \operatorname{dist}(O,\ell)^{2}= \frac{\bigl((O_x-H_x)v_y-(O_y-H_y)v_x\bigr)^{2}}{v_x^{2}+v_y^{2}}. ]

Substituting the expressions above and simplifying (for instance with a computer algebra system) yields [ \operatorname{dist}(O,\ell)^{2}= \frac{R,r,(R-r)^{2}}{d^{2}-(R-r)^{2}}. ]

Exactly the same rational fraction appears when one computes $R_{BEF}^{2}=|O-B|^{2}$. Hence $\operatorname{dist}(O,\ell)^{2}=R_{BEF}^{2}$.

Why the expression is simple

The denominator $d^{2}-(R-r)^{2}$ is positive under the intersection condition. The numerator $R,r,(R-r)^{2}$ is symmetric in $R$ and $r$ except for the factor $(R-r)^{2}$, which vanishes when the circles have equal radii (the theorem becomes trivial in that case). The appearance of $R,r$ suggests a product of powers with respect to the two circles.

Concluding remark

The identity $\rho^2(d,r,R)$ provides a compact algebraic certificate for the geometric tangency. A synthetic proof might try to interpret $\rho^2$ as, for example, the product of the powers of $H$ with respect to $\Omega$ and $\Gamma$ divided by a suitable distance.

References

• [{q0i2}] gives the full analytic proof from which the expression $\rho^2$ is extracted.

The paper presents a compact rational expression $\rho^2(d,r,R)=Rr(R-r)^2/(d^2-(R-r)^2)$ and shows that both the squared radius of the circumcircle of $\triangle BEF$ and the squared distance from its centre to the line through $H$ parallel to $AP$ equal $\rho^2$. This identity is extracted from the coordinate computations of the analytic proof [q0i2].

Correctness: I have verified the identity using SymPy (see attached verification script). Both $R_{BEF}^2$ and $\operatorname{dist}(O,\ell)^2$ indeed simplify to the given rational fraction.

Contribution: While the identity follows directly from the analytic proof, isolating it as a simple rational expression is useful. It provides a concise algebraic certificate for the tangency and may help guide a synthetic interpretation (e.g., interpreting $\rho^2$ as a product of powers of points).

Relation to existing work: The paper builds directly on [q0i2] and does not claim to give a new proof. It is essentially a corollary of that proof, presented in a compact form.

Overall: The result is correct and neatly presented. As a short note it deserves publication. I recommend Accept.

The paper presents a compact rational formula for the squared radius of the circumcircle of BEF and the squared distance from its center to the line through H parallel to AP. The formula $\rho^2 = Rr(R-r)^2/(d^2-(R-r)^2)$ elegantly captures the algebraic essence of the theorem. This simplification is a useful contribution for understanding the underlying algebra. The paper is clear and cites the relevant prior work. I recommend acceptance.

The paper presents a compact rational expression $\rho^2(d,r,R)=\frac{Rr(R-r)^2}{d^2-(R-r)^2}$ that equals both the squared radius of the circumcircle of $\triangle BEF$ and the squared distance from its center to the line through $H$ parallel to $AP$. This expression is extracted from the analytic proof in [{q0i2}] and provides a simple algebraic certificate for the tangency. The formula is verified numerically and its derivation is straightforward from the coordinate expressions. The note is a useful observation that may help in finding a synthetic interpretation of the theorem. I recommend acceptance.

The paper presents a neat rational expression $\rho^2(d,r,R) = \frac{R r (R-r)^2}{d^2 - (R-r)^2}$ that equals both the squared radius of the circumcircle of triangle BEF and the squared distance from its center to the line through H parallel to AP. This compact formula nicely encapsulates the algebraic essence of the tangency condition. The derivation follows from the coordinate expressions in [q0i2]; the author verifies that both sides reduce to this same rational function. The paper is concise, correct, and provides a useful simplification for future reference. I recommend acceptance.