Apollonius circle

theorem

The locus of points whose distances to two fixed points have a constant ratio k≠1 is a circle; commonly used to construct an auxiliary point that converts "weighted-distance extrema" into plain-distance form.

Apollonius circle — a similarity tool that converts "weighted distance" into "unit-weight distance".

Universal form

min(PA+kPB),0<k<1\min\big(\, PA + k \cdot PB \,\big), \quad 0 < k < 1

Classical model

  • Moving point PP: on a circle C\odot C (radius rr)
  • Fixed point AA: anywhere outside the circle
  • Fixed point BB: outside the circle, distinct from the centre CC
  • Key constraint: k=rCBk = \dfrac{r}{CB} (if not, rescale the figure first via similarity; otherwise this is not an Apollonius-circle problem)

Moving point P on circle ⊙C (sliding along the circumference) + outside fixed points A, B; key constraint k = r/CB

Construction and proof

  1. On ray CBCB, take an auxiliary point BB' with

    CB=r2CBCBCB=r2=CP2CB' = \dfrac{r^2}{CB} \quad\Leftrightarrow\quad CB' \cdot CB = r^2 = CP^2

    Construction of B': on ray CB take CB' = r²/CB, equivalent to CB·CB' = CP²

  2. By CBCP=CPCB=k\dfrac{CB'}{CP} = \dfrac{CP}{CB} = k and the shared angle BCP\angle BCP,

    CPBCBP(SAS)\triangle CPB' \sim \triangle CBP \quad (\text{SAS})

    SAS similarity △CPB' ~ △CBP: CB'/CP = CP/CB = k, shared angle ∠BCP

  3. The similarity ratio gives PBPB=k\dfrac{PB'}{PB} = k, so

    PB=kPBPB' = k \cdot PB

    for every PP on the circle — this is the key invariant.

  4. Therefore PA+kPB=PA+PBABPA + k \cdot PB = PA + PB' \geq |AB'| ([[triangle-inequality]]), with equality when A,P,BA, P, B' are collinear and PP lies between AA and BB'.

    When A, P, B' are collinear, PA + k·PB = |AB'| attains the minimum

Apollonius construction: P on ⊙C, A outside, auxiliary B' (CB·CB'=r²) makes PB' = k·PB

Answer

min(PA+kPB)=AB\min\big(\, PA + k \cdot PB \,\big) = |AB'|

where AB|AB'| is computed directly from the geometry of ACB\triangle ACB'.

Applications

  • [[0003-apollonius-circle]] — the standard right-triangle-plus-circle problem