8.(2025·浙江模拟)如图,$\odot O$的直径$AB=10$,将$\odot O$沿$CD$折叠,使$\overset{\frown}{CED}$与直径$AB$相切于点$E$,则折痕$CD$的取值范围为 （）

A.$5≤ CD≤ 5\sqrt{2}$
B.$5≤ CD≤ 5\sqrt{3}$
C.$5\sqrt{2}≤ CD≤ 5\sqrt{3}$
D.$5＜ CD＜ 5\sqrt{3}$

8.C

9.(2025·湖州一模)如图，$\odot O$的半径是1，PA，PB分别切$\odot O$于点A，B，连接PO并延长交$\odot O$于点C，连接AC，BC.若四边形PACB是菱形，则PC的长是.

9.3

10.在平面直角坐标系$xOy$中，$A$为$y$轴正半轴上一点，点$B(1,0)$，$C(5,0)$，$\odot P$是$△ ABC$的外接圆.当$∠ BAC$最大时，点$A$的坐标为.

10.$(0,\sqrt{5})$

11.(2025·镇江模拟)如图,AB为$\odot O$的直径,点C在直径AB上(点C与A,B两点不重合),$OC=3$,点D在$\odot O$上且满足$AC=AD$,连接DC并延长到点E,使$BE=BD$.
(1)求证:BE是$\odot O$的切线;
(2)当$BE=6$时,求$\odot O$的半径.

11.(1)证明:$\because AB$为$\odot O$的直径,$\therefore ∠ADB=90^{\circ },$
$\therefore ∠BDE+∠ADC=90^{\circ }.$
$\because AC=AD,\therefore ∠ACD=∠ADC.$
$\because ∠ACD=∠ECB,\therefore ∠ECB=∠ADC.$
$\because EB=DB,\therefore ∠E=∠BDE,$
$\therefore ∠E+∠BCE=90^{\circ },$
$\therefore ∠EBC=180^{\circ }-(∠E+∠ECB)=90^{\circ }.$
$\because OB$是$\odot O$的半径,$\therefore BE$是$\odot O$的切线.
(2)解:设$\odot O$的半径为$r$,$\because OC=3,$
$\therefore AC=AD=AO+OC=3+r.$
$\because BE=6,\therefore BD=BE=6.$
在$\mathrm{Rt}△ABD$中,$BD^{2}+AD^{2}=AB^{2},$
$\therefore 6^{2}+(r+3)^{2}=(2r)^{2},$
解得$r_1=5,r_2=-3$(舍去),$\therefore \odot O$的半径为5.

12.如图，$\odot O$的半径为5，B，C是$\odot O$上两定点，A是$\odot O$上一动点，且$∠ BAC=60°$，$∠ BAC$的平分线交$\odot O$于点D.
(1)求证：D为$\overset{\frown}{BC}$上一定点.
(2)过点D作BC的平行线交AB的延长线于点F.
①判断DF与$\odot O$的位置关系，并说明理由；
②若$△ ABC$为锐角三角形，求DF的取值范围.

12.(1)证明:连接 OB,OD,如答图①.
$\because ∠BAC=60^{\circ },∠BAC$的平分线交$\odot O$于点 D,
$\therefore ∠BAD=\frac{1}{2}∠BAC=30^{\circ },\therefore ∠BOD=2∠BAD=60^{\circ },$
$\therefore \overset{\frown}{BD}$的度数是$60^{\circ }.$
$\because B$为定点,$\therefore D$为$\overset{\frown}{BC}$上一定点.
(2)解:①DF 与$\odot O$相切.理由如下:
连接 OD,如答图②.
$\because ∠BAC$的平分线交$\odot O$于点 D,
$\therefore ∠BAD=∠CAD,\therefore \overset{\frown}{BD}=\overset{\frown}{CD},\therefore OD⊥BC.$
$\because DF// BC,\therefore OD⊥DF.$
$\because OD$为$\odot O$的半径,$\therefore DF$与$\odot O$相切.
②当$∠A_1BC$为直角时,连接 OD 交 BC 于点 M,如答图③.
$\because ∠BA_1C=60^{\circ },∠A_1BC=90^{\circ },$
$\therefore ∠C=30^{\circ },A_1C$为$\odot O$的直径.
$\because \odot O$的半径为5,$\therefore A_1C=10,A_1B=\frac{1}{2}A_1C=5,$
$\therefore BC=5\sqrt{3}$,由①知$\overset{\frown}{BD}=\overset{\frown}{CD},$
$\therefore BM=\frac{1}{2}BC=\frac{5\sqrt{3}}{2},∠BMD=90^{\circ }.$
$\because ∠FBC=180^{\circ }-∠A_1BC=90^{\circ },∠FDM=90^{\circ },$
$\therefore$四边形 BFDM 是矩形,$\therefore DF=BM=\frac{5\sqrt{3}}{2};$
当$∠A_2CB$为直角时,连接 OD,BD,如答图④.
$\because ∠A_2CB=90^{\circ },∠BA_2C=60^{\circ },$
$\therefore A_2B$是$\odot O$的直径,$∠A_2BC=30^{\circ }.$
$\because DF// BC,\therefore ∠F=∠A_2BC=30^{\circ }.$
$\because DF$与$\odot O$相切,$\therefore ∠FDO=90^{\circ },$
$\therefore OF=2OD=10,$
$\therefore DF=\sqrt{OF^2-OD^2}=\sqrt{10^2-5^2}=5\sqrt{3}.$
由图可知,当点 A 由点$A_1$运动到点$A_2$(不包括点$A_1$,点$A_2$)时,$△ABC$是锐角三角形,
$\therefore DF$的取值范围是$\frac{5\sqrt{3}}{2}＜DF＜5\sqrt{3}.$