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1. [2025南京栖霞区八校联考月考]如图，已知PA，PB分别切$\odot O$于点A，B，$∠ P=60°$，$PA=8$，则弦AB的长是（
A.4
B.8
C.$4\sqrt{3}$
D.$8\sqrt{3}$
B
）A.4
B.8
C.$4\sqrt{3}$
D.$8\sqrt{3}$
答案
∵ PA,PB分别切$\odot O$于点A,B,$\therefore PA = PB$. 又$\because ∠ P = 60°,\therefore △ APB$是等边三角形,$\therefore AB = PA = 8$.
2.［2026常州市北实验中学期中］如图，P为⊙O外一点，PA，PB分别切⊙O于点A，B，CD切⊙O于点E，PA，PB分别与CD交于点C，D。若PA=10，则△PCD的周长为（
A.10
B.12
C.16
D.20
D
）A.10
B.12
C.16
D.20
答案
$\because$ PA,PB分别切$\odot O$于点A,B,CD切$\odot O$于点E,$\therefore AC = CE,DE = DB,PA = PB,\therefore △ PCD$的周长$= PC + CD + PD = PC + CE + DE + PD = PA + PB = 2PA. \because PA = 10,$$\therefore △ PCD$的周长为20.
3. 如图,正方形ABCD的边长为4 cm,以正方形的一边BC为直径,在正方形ABCD内作半圆O,再过点A作半圆O的切线,与半圆O相切于点F,与DC相交于点E,则△ADE的面积为 (
A.5 cm²
B.6 cm²
C.7 cm²
D.8 cm²
B
)A.5 cm²
B.6 cm²
C.7 cm²
D.8 cm²
答案
$\because$ 四边形ABCD是正方形,$\therefore ∠ ABC = 90°,\therefore AB ⊥ OB,\therefore AB$是半圆O的切线,切点为B.同理可证CD与半圆O相切于点C. $\because AE$与$\odot O$相切于点F,$\therefore AF = AB = 4\ \mathrm{cm},EF = EC$. 设$EF = EC = x\ \mathrm{cm}$,则$DE = (4 - x)\mathrm{cm},AE = (4 + x)\mathrm{cm}$. 在$\mathrm{Rt}△ ADE$中,由勾股定理得$DE^{2} + AD^{2} = AE^{2},\therefore (4 - x)^{2} + 4^{2} = (4 + x)^{2}$,解得$x = 1$,即$CE = 1\ \mathrm{cm},\therefore DE = 4 - 1 = 3(\mathrm{cm})$,$\therefore S_{△ ADE} = \frac{1}{2}AD · DE = \frac{1}{2} × 4 × 3 = 6(\mathrm{cm}^{2})$.
4. [2025 南京金陵中学河西分校期中] 如图，AB，BC，CD，DA都是$\odot O$的切线，$AD=2$，$AB+CD=8$，则$BC=$
6
。答案
如图,设AB,BC,CD,DA与$\odot O$的切点分别为E,H,G,F. $\because$ AB,BC,CD,DA都是$\odot O$的切线,$\therefore AF = AE,BE = BH$,$CH = CG,DG = DF,\therefore AD + BC = AF + DF + BH + CH = AE + DG + BE + GC = AE + BE + DG + CG = AB + CD. \because AD = 2,AB + CD = 8,\therefore BC = AB + CD - AD = 8 - 2 = 6.$
5. 为了测量一个圆形铁环的半径，小明采用如下方法：将铁环平放在水平桌面上，用一个锐角为$30°$的三角板和一把刻度尺，按照如图所示的方法得到相关数据，进而可求得铁环的半径。若测得$PA=1\ \mathrm{cm}$，则铁环的半径是
$\sqrt{3}$
cm.答案
设$\odot O$与CA相切于点N,如图,连接OP,OA,ON.
$\because$ PA,CA与$\odot O$相切,$\therefore AP = AN,∠ OPA = ∠ ONA = 90°$. 在$\mathrm{Rt}△ AOP$和$\mathrm{Rt}△ AON$中,$\begin{cases} OA = OA, \\ AP = AN, \end{cases} \therefore \mathrm{Rt}△ AOP ≌ \mathrm{Rt}△ AON(\mathrm{HL}),$
$\therefore ∠ OAP = ∠ OAN$. 在$\mathrm{Rt}△ ABC$中,$∠ BAC = 90° - 30° = 60°$,
$\therefore ∠ PAN = 180° - 60° = 120°,\therefore ∠ OAP = ∠ OAN = 60°$,
$\therefore ∠ POA = 90° - ∠ OAP = 90° - 60° = 30°$. 又$\because ∠ OPA = 90°$,
$\therefore OA = 2PA = 2 × 1 = 2(\mathrm{cm})$. 在$\mathrm{Rt}△ OPA$中,由勾股定理得
$OP = \sqrt{OA^{2} - AP^{2}} = \sqrt{2^{2} - 1^{2}} = \sqrt{3}(\mathrm{cm})$.
6. 如图,直线AB,BC,CD分别与$\odot O$相切于点E,F,G,且$AB// CD$,$OB=6\ \mathrm{cm}$,$OC=8\ \mathrm{cm}$.求:
(1)$∠ BOC$的度数;
(2)BE和CG的长度之和;
(3)$\odot O$的半径.
(1)$∠ BOC$的度数;
(2)BE和CG的长度之和;
(3)$\odot O$的半径.
答案
解:(1)如图,连接OE,OF,OG.
$\because$ AB,BC分别与$\odot O$相切于点E,F,$\therefore BE = BF,∠ OEB = ∠ OFB = 90°$.
又$\because OB = OB,\therefore \mathrm{Rt}△ BOE ≌ \mathrm{Rt}△ BOF$,
$\therefore ∠ OBF = ∠ OBE$,同理可得$∠ OCF = ∠ OCG$.
$\because AB // CD,\therefore ∠ ABC + ∠ BCD = 180°$,
$\therefore ∠ OBF + ∠ OCF = 90°,\therefore ∠ BOC = 90°$.
(2)由(1)知,$∠ BOC = 90°$.
$\because OB = 6\ \mathrm{cm},OC = 8\ \mathrm{cm}$,
$\therefore$ 由勾股定理,得$BC = \sqrt{OB^{2} + OC^{2}} = 10\ \mathrm{cm}$,
$\therefore BE + CG = BF + CF = BC = 10\ \mathrm{cm}$.
(3)由(1)知$OF ⊥ BC,OB ⊥ OC$,
$\therefore S_{△ OBC} = \frac{1}{2}OF × BC = \frac{1}{2}OB × OC$,
即$\frac{1}{2}OF × 10 = \frac{1}{2} × 6 × 8$,
$\therefore OF = 4.8\ \mathrm{cm},\therefore \odot O$的半径为4.8 cm.
7. 如图，AB为$\odot O$的直径，过圆外一点E作$\odot O$的两条切线EC，EB，切点分别为点D，B，EC交BA的延长线于点C，连接OE，AD.
(1)AD与OE有怎样的位置关系？并说明理由.
(2)若$EB=6$，$CD=4$，求$\odot O$的半径.
(1)AD与OE有怎样的位置关系？并说明理由.
(2)若$EB=6$，$CD=4$，求$\odot O$的半径.
答案
解:(1)$AD // OE$. 理由如下:
连接OD.
$\because CE,BE$是$\odot O$的切线,
$\therefore ∠ ODE = ∠ OBE = 90°$.
在$\mathrm{Rt}△ DOE$和$\mathrm{Rt}△ BOE$中,$\begin{cases} OE = OE, \\ OD = OB, \end{cases}$
$\therefore \mathrm{Rt}△ DOE ≌ \mathrm{Rt}△ BOE(\mathrm{HL})$.
$\therefore ∠ DOE = ∠ BOE$.
$\therefore ∠ DOB = ∠ DOE + ∠ BOE = 2∠ BOE$.
$\because OA = OD,\therefore ∠ ODA = ∠ OAD$.
$\because ∠ DOB = ∠ ODA + ∠ OAD$,
$\therefore ∠ DOB = 2∠ OAD$.
$\therefore ∠ BOE = ∠ OAD,\therefore AD // OE$.
(2)$\because CE,BE$是$\odot O$的切线,
$\therefore DE = BE = 6,\therefore CE = CD + DE = 6 + 4 = 10$.
在$\mathrm{Rt}△ BCE$中,由勾股定理得 $BC = \sqrt{CE^{2} - BE^{2}} = \sqrt{10^{2} - 6^{2}} = 8$.
设$OB = OD = r$,则$OC = 8 - r$.
在$\mathrm{Rt}△ OCD$中,由勾股定理得$CD^{2} + OD^{2} = OC^{2}$,
$\therefore 4^{2} + r^{2} = (8 - r)^{2}$,解得$r = 3$,
即$\odot O$的半径为3.
连接OD.
$\because CE,BE$是$\odot O$的切线,
$\therefore ∠ ODE = ∠ OBE = 90°$.
在$\mathrm{Rt}△ DOE$和$\mathrm{Rt}△ BOE$中,$\begin{cases} OE = OE, \\ OD = OB, \end{cases}$
$\therefore \mathrm{Rt}△ DOE ≌ \mathrm{Rt}△ BOE(\mathrm{HL})$.
$\therefore ∠ DOE = ∠ BOE$.
$\therefore ∠ DOB = ∠ DOE + ∠ BOE = 2∠ BOE$.
$\because OA = OD,\therefore ∠ ODA = ∠ OAD$.
$\because ∠ DOB = ∠ ODA + ∠ OAD$,
$\therefore ∠ DOB = 2∠ OAD$.
$\therefore ∠ BOE = ∠ OAD,\therefore AD // OE$.
(2)$\because CE,BE$是$\odot O$的切线,
$\therefore DE = BE = 6,\therefore CE = CD + DE = 6 + 4 = 10$.
在$\mathrm{Rt}△ BCE$中,由勾股定理得 $BC = \sqrt{CE^{2} - BE^{2}} = \sqrt{10^{2} - 6^{2}} = 8$.
设$OB = OD = r$,则$OC = 8 - r$.
在$\mathrm{Rt}△ OCD$中,由勾股定理得$CD^{2} + OD^{2} = OC^{2}$,
$\therefore 4^{2} + r^{2} = (8 - r)^{2}$,解得$r = 3$,
即$\odot O$的半径为3.
