Proof using vectors in Euclidean spaces

Let d₁, d₂ be non-equal positive real numbers. Let C be the internal division point of AB in the ratio d₁ : d₂ and D the external division point of AB in the same ratio, d₁ : d₂.

${\displaystyle {\overrightarrow {\mathrm {PC} }}={\frac {d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}+d_{1}}},\ {\overrightarrow {\mathrm {PD} }}={\frac {d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}-d_{1}}}.}$

Then,

${\displaystyle {\begin{aligned}&\mathrm {PA}$ :\mathrm {PB} =d_{1}:d_{2}.\\\Leftrightarrow {}&d_{2}|{\overrightarrow {\mathrm {PA} }}|=d_{1}|{\overrightarrow {\mathrm {PB} }}|.\\\Leftrightarrow {}&d_{2}^{2}|{\overrightarrow {\mathrm {PA} }}|^{2}=d_{1}^{2}|{\overrightarrow {\mathrm {PB} }}|^{2}.\\\Leftrightarrow {}&(d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }})\cdot (d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }})=0.\\\Leftrightarrow {}&{\frac {d_{2}{\overrightarrow {\mathrm {PA} }}+d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}+d_{1}}}\cdot {\frac {d_{2}{\overrightarrow {\mathrm {PA} }}-d_{1}{\overrightarrow {\mathrm {PB} }}}{d_{2}-d_{1}}}=0.\\\Leftrightarrow {}&{\overrightarrow {\mathrm {PC} }}\cdot {\overrightarrow {\mathrm {PD} }}=0.\\\Leftrightarrow {}&{\overrightarrow {\mathrm {PC} }}={\vec {0}}\vee {\overrightarrow {\mathrm {PD} }}={\vec {0}}\vee {\overrightarrow {\mathrm {PC} }}\perp {\overrightarrow {\mathrm {PD} }}.\\\Leftrightarrow {}&\mathrm {P} =\mathrm {C} \vee \mathrm {P} =\mathrm {D} \vee \angle {\mathrm {CPD} }=90^{\circ }.\end{aligned}}}

Therefore, the point P is on the circle which has the diameter CD.

Proof using the angle bisector theorem

Proof of Apollonius' definition of a circle

First consider the point ${\displaystyle C}$ on the line segment between ${\displaystyle A}$ and ${\displaystyle B}$, satisfying the ratio. By the definition

$${\displaystyle {\frac {|AP|}{|BP|}}={\frac {|AC|}{|BC|}}}$$

and from the angle bisector theorem the angles ${\displaystyle \alpha =\angle APC}$ and ${\displaystyle \beta =\angle CPB}$ are equal.

Next take the other point ${\displaystyle D}$ on the extended line ${\displaystyle AB}$ that satisfies the ratio. So

$${\displaystyle {\frac {|AP|}{|BP|}}={\frac {|AD|}{|BD|}}.}$$

Also take some other point ${\displaystyle Q}$ anywhere on the extended line ${\displaystyle AP}$. Also by the Angle bisector theorem the line ${\displaystyle PD}$ bisects the exterior angle ${\displaystyle \angle QPB}$. Hence, ${\displaystyle \gamma =\angle BPD}$ and ${\displaystyle \delta =\angle QPD}$ are equal and ${\displaystyle \beta +\gamma =90^{\circ }}$. Hence by Thales's theorem ${\displaystyle P}$ lies on the circle which has ${\displaystyle CD}$ as a diameter.

Apollonius pursuit problem

The Apollonius pursuit problem is one of finding whether a ship leaving from one point A at speed v_A will intercept another ship leaving a different point B at speed v_B. The minimum time in interception of the two ships is calculated by means of straight-line paths. If the ships' speeds are held constant, their speed ratio is defined by μ. If both ships collide or meet at a future point, I, then the distances of each are related by the equation:^[1]

${\displaystyle a=\mu b}$

Squaring both sides, we obtain:

${\displaystyle a^{2}=b^{2}\mu ^{2}}$

${\displaystyle a^{2}=x^{2}+y^{2}}$

${\displaystyle b^{2}=(d-x)^{2}+y^{2}}$

${\displaystyle x^{2}+y^{2}=[(d-x)^{2}+y^{2}]\mu ^{2}}$

Expanding:

${\displaystyle x^{2}+y^{2}=[d^{2}+x^{2}-2dx+y^{2}]\mu ^{2}}$

Further expansion:

${\displaystyle x^{2}+y^{2}=x^{2}\mu ^{2}+y^{2}\mu ^{2}+d^{2}\mu ^{2}-2dx\mu ^{2}}$

Bringing to the left-hand side:

${\displaystyle x^{2}-x^{2}\mu ^{2}+y^{2}-y^{2}\mu ^{2}-d^{2}\mu ^{2}+2dx\mu ^{2}=0}$

Factoring:

${\displaystyle x^{2}(1-\mu ^{2})+y^{2}(1-\mu ^{2})-d^{2}\mu ^{2}+2dx\mu ^{2}=0}$

Dividing by ${\displaystyle 1-\mu ^{2}}$ :

${\displaystyle x^{2}+y^{2}-{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}+{\frac {2dx\mu ^{2}}{1-\mu ^{2}}}=0}$

Completing the square:

${\displaystyle \left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}-{\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}-{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}+y^{2}=0}$

Bring non-squared terms to the right-hand side:

${\displaystyle {\begin{aligned}\left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}+y^{2}&={\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}+{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}\\&={\frac {d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}+{\frac {d^{2}\mu ^{2}}{1-\mu ^{2}}}{\frac {(1-\mu ^{2})}{(1-\mu ^{2})}}\\&={\frac {d^{2}\mu ^{4}+d^{2}\mu ^{2}-d^{2}\mu ^{4}}{(1-\mu ^{2})^{2}}}\\&={\frac {d^{2}\mu ^{2}}{(1-\mu ^{2})^{2}}}\end{aligned}}}$

Then:

${\displaystyle \left(x+{\frac {d\mu ^{2}}{1-\mu ^{2}}}\right)^{2}+y^{2}=\left({\frac {d\mu }{1-\mu ^{2}}}\right)^{2}}$

Therefore, the point must lie on a circle as defined by Apollonius, with their starting points as the foci.