アポロニウスの円
線分QRの中点をOとすると、
O Q → = − 1 2 Q R → , O R → = 1 2 Q R → . {\displaystyle {\overrightarrow {\mathrm {OQ} }}=-{\frac {1}{2}}{\overrightarrow {\mathrm {QR} }},\ {\overrightarrow {\mathrm {OR} }}={\frac {1}{2}}{\overrightarrow {\mathrm {QR} }}.}
したがって、
P Q → ⋅ P R → = 0. {\displaystyle {\overrightarrow {\mathrm {PQ} }}\cdot {\overrightarrow {\mathrm {PR} }}=0.}
⇔ ( P O → + O Q → ) ⋅ ( P O → + O R → ) = 0. {\displaystyle \Leftrightarrow ({\overrightarrow {\mathrm {PO} }}+{\overrightarrow {\mathrm {OQ} }})\cdot ({\overrightarrow {\mathrm {PO} }}+{\overrightarrow {\mathrm {OR} }})=0.}
⇔ ( P O → − 1 2 Q R → ) ⋅ ( P O → + 1 2 Q R → ) = 0.