典例1（2023·宿迁）在同一平面内，已知⊙O的半径为2，圆心O到直线l的距离为3，P为圆上的一个动点，则点P到直线l的最大距离是（ ）
A. 2
B. 5
C. 6
D. 8

B.

典例2（2024·苏州虎丘模拟）已知P是半径为4的⊙O上一点，平面上一点Q到点P的距离为2，则线段OQ的长度a的取值范围是____________.

$2\leqslant a\leqslant 6$.

典例3（2023·无锡）如图，AB是⊙O的切线，B为切点，OA与BC交于点D，AB = AD.若∠C = 20°，则∠OAB的度数为（ ）

A. 20°
B. 30°
C. 40°
D. 50°

C.

典例4（2024·扬州广陵一模）如图，AB为⊙O的直径，点C在⊙O上，∠ACB的平分线交⊙O于点D，过点D作DE//AB，交CB的延长线于点E，连接AD，BD.
(1)求证：DE是⊙O的切线；
(2)若AC = 3$\sqrt{2}$，BC = $\sqrt{2}$，求BD，CD的长.

（1）如图，连接OD. $\because CD$是$\angle ACB$的平分线，$\therefore \angle ACD=\angle BCD$. $\therefore \angle AOD=\angle BOD$. $\because AB$为$\odot O$的直径，$\therefore \angle AOD=\angle BOD=\frac{1}{2}\times 180^{\circ}=90^{\circ}$. $\therefore OD\perp AB$. $\because DE// AB$，$\therefore OD\perp DE$. $\because OD$为$\odot O$的半径，$\therefore DE$是$\odot O$的切线.
（2）$\because AB$为$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}$，$\angle ADB = 90^{\circ}$.
$\because AC = 3\sqrt{2}$，$BC = \sqrt{2}$，$\therefore AB = \sqrt{AC^{2}+BC^{2}} = 2\sqrt{5}$.
$\because \angle ACB$的平分线$CD$交$\odot O$于点$D$，$\therefore \angle ACD=\angle BCD$.
$\therefore \overset{\frown}{AD}=\overset{\frown}{BD}$. $\therefore AD = BD=\frac{\sqrt{2}}{2}AB = \sqrt{10}$. 如图，过点$B$作$BH\perp CD$于点$H$. $\because \angle BCD=\frac{1}{2}\angle ACB = 45^{\circ}$，$\therefore BH = CH=\frac{\sqrt{2}}{2}BC = 1$. $\therefore DH = \sqrt{BD^{2}-BH^{2}} = 3$. $\therefore CD = CH + DH = 1 + 3 = 4$.

[变式]（2024·徐州铜山二模）如图，在⊙O中，AB是直径，点C在⊙O上，点D在AB的延长线上，连接CD，使∠BCD = ∠A.
（1）求证：直线CD是⊙O的切线；
（2）若AC = CD，BD = 2，求AB的长.

（1）如图，连接$OC$，则$OB = OC$. $\therefore \angle OBC=\angle OCB$.
$\because AB$是$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}$. $\therefore \angle A+\angle ABC = 90^{\circ}$.
$\because \angle BCD=\angle A$，$\therefore \angle BCD+\angle OCB = 90^{\circ}$，即$\angle OCD = 90^{\circ}$.
$\therefore OC\perp CD$. $\because OC$是$\odot O$的半径，$\therefore$直线$CD$是$\odot O$的切线.
（2）$\because OA = OC$，$\therefore \angle A = \angle OCA$. $\therefore \angle BOC = \angle A+\angle OCA = 2\angle A$. $\because AC = CD$，$\therefore \angle A=\angle D$. $\because \angle BCD=\angle A$，$\therefore \angle BCD=\angle D$. $\therefore \angle OBC=\angle BCD+\angle D = 2\angle D = 2\angle A$.
$\therefore \angle OBC=\angle BOC$. $\therefore OC = BC$. $\because OB = OC$，$\therefore OB = OC = BC$. $\therefore \triangle OBC$是等边三角形. $\therefore \angle COD = 60^{\circ}$. 由（1）知，$\angle OCD = 90^{\circ}$，$\therefore \angle D = 30^{\circ}$. $\therefore OC=\frac{1}{2}OD$. $\therefore OD = 2OC = 2OB$. $\because OD = OB + BD$，$\therefore OB = BD = 2$. $\therefore AB = 2OB = 4$.

典例5（2023·镇江）《九章算术》中记载：“今有勾八步，股一十五步.问：勾中容圆径几何？”
译文：今有一个直角三角形，勾（短直角边）长为8步，股（长直角边）长为15步，问：该直角三角形内切圆的直径是多少？书中给出的算法
译文如下：如图，根据勾、股，求得弦长. 用勾、股、弦相加作为除数，用勾乘股，再乘2作为被除数，商即为该直角三角形内切圆的直径，求得该直径为________步（注：“步”为长度单位）.

6.