A circle is usually defined as the set of points
P at a given distance
r (the circle's radius) from a given point (the circle's center). However, there are other, equivalent definitions of a circle. Apollonius discovered that a circle could be defined as the set of points
P that have a given
ratio of distances
k = to two given points (labeled
A and
B in the figure). These two points are sometimes called the
foci.
Proof using vectors in Euclidean spaces Let
d,
d be non-equal positive real numbers. Let
C be the internal division point of
AB in the ratio
d :
d and
D the external division point of
AB in the same ratio,
d :
d. :\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, :\begin{align} &\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{align} Therefore, the point
P is on the circle which has the diameter
CD.
Proof using the angle bisector theorem First consider the point C on the line segment between A and B, satisfying the ratio. By the definition \frac=\frac and from the converse of the
angle bisector theorem, the angles \alpha=\angle APC and \beta=\angle CPB are equal. Next take the other point D on the extended line AB that satisfies the ratio. So \frac=\frac. Also take some other point Q anywhere on the extended line AP. Again by the converse of the angle bisector theorem, the line PD bisects the exterior angle \angle QPB. Hence, \gamma=\angle BPD and \delta=\angle QPD are equal and \beta+\gamma=90^{\circ}. Hence by
Thales's theorem P lies on the circle which has 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
vA will intercept another ship leaving a different point
B at speed
vB. 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: :a = \mu b Squaring both sides, we obtain: :a^{2} = b^{2} \mu^{2} :a^{2} = x^{2} + y^{2} :b^{2} = (d-x)^{2} + y^{2} :x^{2} + y^{2} = [(d-x)^{2} + y^{2}]\mu^{2} Expanding: :x^{2}+y^{2} = [d^{2} + x^{2} - 2dx + y^{2}]\mu^{2} Further expansion: :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: :x^{2} - x^{2}\mu^{2} + y^{2} - y^{2}\mu^{2} - d^{2}\mu^{2} + 2dx\mu^{2} = 0 Factoring: :x^{2}(1-\mu^{2}) + y^{2}(1-\mu^{2}) - d^{2}\mu^{2} + 2dx\mu^{2} = 0 Dividing by 1-\mu^{2} : :x^{2} + y^{2} - \frac{d^{2}\mu^{2}}{1-\mu^{2}} + \frac{2dx\mu^{2}}{1-\mu^{2}} = 0 Completing the square: :\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: :\begin{align} \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{align} Then: :\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. ==Circles sharing a radical axis==