1.如圖,點B在$\odot A$上,點C在$\odot A$外,以下條件不能判定BC是$\odot A$的切線的是(
D
)
A.$∠A= 50^{\circ },∠C= 40^{\circ }$
B.$∠B-∠C= ∠A$
C.$AB^{2}+BC^{2}= AC^{2}$
D.$\odot A$與AC的交點是AC的中點
答案：D
A. $\because ∠A=50^{\circ },∠C=40^{\circ },\therefore ∠B=180^{\circ }-∠A-∠C=90^{\circ },\therefore BC⊥AB$,∵點B在$\odot A$上,$\therefore AB$是$\odot A$的半徑,$\therefore BC$是$\odot A$的切線;
B. $\because ∠B-∠C=∠A,\therefore ∠B=∠A+∠C,\because ∠A+∠B+∠C=180^{\circ },\therefore ∠B=90^{\circ },\therefore BC⊥AB$,∵點B在$\odot A$上,$\therefore AB$是$\odot A$的半徑,$\therefore BC$是$\odot A$的切線;
C. $\because AB^{2}+BC^{2}=AC^{2},\therefore △ABC$是直角三角形,$∠B=90^{\circ },\therefore BC⊥AB$,∵點B在$\odot A$上,$\therefore AB$是$\odot A$的半徑,$\therefore BC$是$\odot A$的切線;
D. $\because \odot A$與AC的交點是AC的中點,$\therefore AB=\frac {1}{2}AC$,但不能證出$∠B=90^{\circ },\therefore $不能判定BC是$\odot A$的切線.故選D.
2.學科教材變式如圖,$OA= OB= 13cm,AB= 24cm,\odot O$的直徑為10cm.求證：AB是$\odot O$的切線.
答案：證明 如圖,過點O作$OC⊥AB$于點C,
$\because OA=OB=13cm,AB=24cm,\therefore AC=\frac {1}{2}AB=12cm$.在$Rt△OAC$中,根據(jù)勾股定理,得$OC=\sqrt {OA^{2}-AC^{2}}=5cm$.$\because \odot O$的直徑為10 cm,$\therefore \odot O$的半徑為5 cm,$\therefore OC$為$\odot O$的半徑,$\because OC⊥AB,\therefore AB$是$\odot O$的切線.
3.「2025廣東廣州白云期末」如圖,$\odot O$的直徑AB的長為12,P是AB延長線上的一點,且$PB= 4$,C是$\odot O$上的一點,$PC= 8$.求證：PC是$\odot O$的切線.
答案：證明 如圖，連接OC,
$\because \odot O$的直徑AB的長為12,$\therefore OB=OC=6$,
$\because PB=4,\therefore PO=PB+OB=4+6=10$,
$\because$ 在$△POC$中,$PC=8,OC=6,OP=10$,
$\therefore PC^{2}+OC^{2}=OP^{2},\therefore △POC$是直角三角形,$∠OCP=90^{\circ },\therefore OC⊥PC$,
$\because OC$是$\odot O$的半徑,$\therefore PC$是$\odot O$的切線.
4.「2024山西中考」如圖,已知$\triangle ABC$,以AB為直徑的$\odot O$交BC于點D,與AC相切于點A,連接OD.若$∠AOD= 80^{\circ }$,則$∠C$的度數(shù)為(
D
)
A.$30^{\circ }$
B.$40^{\circ }$
C.$45^{\circ }$
D.$50^{\circ }$
答案：D
$\because \widehat {AD}=\widehat {AD},\therefore ∠B=\frac {1}{2}∠AOD=40^{\circ }.\because $以AB為直徑的$\odot O$與AC相切于點A,$\therefore ∠BAC=90^{\circ },\therefore ∠C=90^{\circ }-40^{\circ }=50^{\circ }$.故選D.
5.「2025江蘇南通海安月考」如圖,$Rt\triangle ABC$中,$∠C= 90^{\circ },BC= 5$,點O在AB上,$OB= 3$,以OB為半徑的$\odot O$與AC相切于點D,交BC于點E,則弦$BE= $____.
答案：答案 4
解析 如圖,連接OD,過點O作$OF⊥BC$于F,$\because AC$與$\odot O$相切于點D,$\therefore OD⊥AC,\because ∠C=90^{\circ },\therefore $四邊形DOFC為矩形,$\therefore FC=OD=3,\therefore BF=BC-FC=5-3=2$,$\because OF⊥BE,\therefore BE=2BF=4$.
6.「2025黑龍江齊齊哈爾龍沙期中」如圖,AB是$\odot O$的直徑,點D是AB延長線上的一點,DC與$\odot O$相切于點C.連接BC,AC.
(1)求證：$∠A= ∠BCD$.
(2)若$∠D= 45^{\circ },\odot O$的半徑為2,求線段AD的長.
答案：解析 (1)證明:如圖,連接OC,$\because DC$是$\odot O$的切線,$\therefore ∠OCD=90^{\circ }$,即$∠BCD+∠OCB=90^{\circ }$,
$\because AB$是$\odot O$的直徑,$\therefore ∠ACB=90^{\circ }$,
$\therefore ∠A+∠OBC=90^{\circ }$,
$\because OC=OB,\therefore ∠OCB=∠OBC,\therefore ∠A=∠BCD$.
(2)在$Rt△OCD$中,$∠D=45^{\circ },OC=2$,
$\therefore OA=OC=2,OD=\sqrt {2}OC=2\sqrt {2}$,
$\therefore AD=OA+OD=2+2\sqrt {2}$.
7.「2024山東濟寧嘉祥期中,★☆」如圖,在平面直角坐標系中,$\odot M$與x軸相切于點A,與y軸交于點B,C.若圓心M的坐標是$(4,5)$,則弦BC的長度為( )
A.3
B.4
C.5
D.6
答案：D
如圖,連接MA、MB、MC,過點M作$MN⊥y$軸,垂足為N,則$BN=CN=\frac {1}{2}BC,\because \odot M$與x軸相切于點A,$\therefore MA⊥x$軸,∵圓心M的坐標是$(4,5),\therefore MA=5$,
$MN=4,\therefore MB=MA=5$. 在$Rt△BMN$中,$BN=\sqrt {MB^{2}-MN^{2}}=\sqrt {5^{2}-4^{2}}=3,\therefore BC=2BN=6$.故選D.
