Exercise 1320Geometry › Plane geometryOlympiadAdded 5 Nov 2026

Two circles and a line through a common point

#spiral similarity#two intersecting circles#chord length#projection

Statement

Two circles $\Gamma_1$ and $\Gamma_2$ , with centres $O_1,O_2$ and radii $R_1,R_2$ , meet at two points $A$ and $B$ . A line $\ell$ through $A$ meets $\Gamma_1$ again at $P$ and $\Gamma_2$ again at $Q$ .

Prove that $\dfrac{BP}{BQ}=\dfrac{R_1}{R_2}$ for every such line $\ell$ .
Prove that $PQ=2\,\bigl|\overrightarrow{O_1O_2}\cdot u\bigr|$ , where $u$ is a unit vector along $\ell$ .
Deduce that $PQ\le2\,O_1O_2$ , with equality exactly when $\ell\parallel O_1O_2$ .

Hint

Part 1: in each circle, a chord equals $2R\sin$ of the inscribed angle that it subtends. Part 2: the foot of the perpendicular from the centre to a chord is the midpoint of the chord.

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Solution

1. In $\Gamma_1$ , the chord $BP$ subtends the inscribed angle $\angle BAP$ at $A$ , so $BP=2R_1\sin\angle BAP$ . In $\Gamma_2$ , $BQ=2R_2\sin\angle BAQ$ . Since $P,A,Q$ are on $\ell$ , the angles $\angle BAP$ and $\angle BAQ$ are equal or supplementary, so their sines are equal. Hence $\frac{BP}{BQ}=\frac{R_1}{R_2}$ .

2. Take $A$ as origin and $u$ as the direction of $\ell$ . The foot of the perpendicular from $O_1$ to $\ell$ is the midpoint of the chord $AP$ , so $P=2(O_1\cdot u)\,u$ . Similarly $Q=2(O_2\cdot u)\,u$ . Hence
$PQ=2\,\bigl|\overrightarrow{O_1O_2}\cdot u\bigr|.$

3. By Cauchy–Schwarz, $\bigl|\overrightarrow{O_1O_2}\cdot u\bigr|\le O_1O_2$ , with equality iff $u\parallel\overrightarrow{O_1O_2}$ . So $PQ\le2\,O_1O_2$ .

Lesson. For two circles meeting at $A$ , project onto the line through $A$ : midpoints of chords are the feet of perpendiculars from the centres.

Two circles and a line through a common point

Statement

Solution

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