elem.3.10
一个圆与另一个圆相交的交点不超过两个。
本页以“圆与圆相交至多两点”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H; let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E. Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC.
假设圆ABC与圆DEF相交于多于两点，即B、C、F、H。连接BH和BG，并分别取中点K和L。
[III. 1, Por.] Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO.
过K作BH的垂线KC，过L作BG的垂线LM，并延长至A和E。
But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC.
由于在圆ABC中，直线AC平分且垂直于BH，故圆心在AC上；同理，直线NO平分且垂直于BG，圆心也在NO上。AC与NO交于唯一一点P，故P为圆ABC的圆心。
Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible.
同理可证P也是圆DEF的圆心，因此两相交圆有同一圆心，这与第三卷命题5矛盾。