《2025年通城学典非常课课通九年级数学上册人教版江苏专版》

示例10（2023·广西改编）如图，$PA$，$PB$切$\odot O$于点$A$，$B$，连接$AO$并延长，交$PB$的延长线于点$C$.已知$\odot O$的半径为$4$，$OC = 5$，求$AP$的长.

答案： 解：如图，连接$OB$.$\therefore OB = OA = 4$.$\because PA$，$PB$是$\odot O$的切线，$\therefore OA\perp PA$，$OB\perp PB$，且$PA = PB$.在$\mathrm{Rt}\triangle OBC$中，$\because OC = 5$，$OB = 4$，$\therefore BC=\sqrt{5^{2}-4^{2}} = 3$.在$\mathrm{Rt}\triangle PAC$中，设$PA = PB = x$，$AC = OA + OC = 4 + 5 = 9$.$\because PA^{2}+AC^{2}=PC^{2}$，$\therefore x^{2}+9^{2}=(x + 3)^{2}$，解得$x = 12$.$\therefore AP$的长为$12$.