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2.如图24.1.1-9，在$Rt\triangle ABC$中，$\angle C = 90^{\circ}$，$AB = 10$.若以点$C$为圆心，$CB$长为半径的圆恰好经过$AB$的中点$D$，则$AC$的长等于 (
A.$5\sqrt{3}$
B.5
C.$5\sqrt{2}$
D.6
A
)A.$5\sqrt{3}$
B.5
C.$5\sqrt{2}$
D.6
答案
2.A
解析
解：在$Rt\triangle ABC$中，$\angle C = 90^{\circ}$，$AB = 10$，$D$为$AB$中点，
$\therefore CD=\frac{1}{2}AB=5$。
$\because$ 圆$C$以$CB$为半径且过点$D$，
$\therefore CB=CD=5$。
在$Rt\triangle ABC$中，由勾股定理得：
$AC=\sqrt{AB^{2}-CB^{2}}=\sqrt{10^{2}-5^{2}}=\sqrt{75}=5\sqrt{3}$。
A
$\therefore CD=\frac{1}{2}AB=5$。
$\because$ 圆$C$以$CB$为半径且过点$D$，
$\therefore CB=CD=5$。
在$Rt\triangle ABC$中，由勾股定理得：
$AC=\sqrt{AB^{2}-CB^{2}}=\sqrt{10^{2}-5^{2}}=\sqrt{75}=5\sqrt{3}$。
A
$3.$如图$24.1.1-10，$若点$O$为$\odot O$的圆心，则线段$\underline{OA（答案不唯一，半径均可）}$是$\odot $
$ $
$O$的半径；线段$\underline{AB（答案不唯一，弦均可）}$是$\odot $ $ $
$O$的弦，其中最长的弦是$\underline{直径（如AB，假设AB是直径）}；$劣弧是$\underline{\overset{\frown}{AC}（答案不唯一）}，$优弧是$\underline{\overset{\frown}{ABC}（答案不唯一）}；$$AC$是$\underline{弦}. $ 答案
3.OA,OC,OB AC,BC,AB $\overset{\frown}{AC}$ $\overset{\frown}{AB}$,$\overset{\frown}{BC}$ $\overset{\frown}{BAC}$,$\overset{\frown}{ACB}$ 半圆
解析
OA,OC,OB；AC,BC,AB；AB；$\overset{\frown}{AC}$；$\overset{\frown}{ABC}$；弦
$4.$下列说法正确的是$\underline{①③④}（$只填序号$）.$
$①$经过点$P$的圆有无数个；$②$以点$P$为圆心的圆有无数个；$③$半径为$3cm$且经过点$P$的圆有无数个；$④$以点$P$为圆心，$3cm$为半径的圆有无数个$.$
$①$经过点$P$的圆有无数个；$②$以点$P$为圆心的圆有无数个；$③$半径为$3cm$且经过点$P$的圆有无数个；$④$以点$P$为圆心，$3cm$为半径的圆有无数个$.$
$①②③$
答案
4.①②③
5.如图24.1.1-11，点$A$，$B$，$C$是$\odot O$上的三点，$BO$平分$\angle ABC$，求证：$BA = BC$.
答案
5.证明:连接OA,OC,如答图24.1.1−3.
因为OA=OB,OB=OC,
所以∠ABO=∠BAO,∠CBO=∠BCO.
因为BO平分∠ABC,
所以∠ABO=∠CBO,
所以∠BAO=∠BCO.因为OB=OB,所以△OAB≌△OCB,所以BA=BC.
6.点$P$到圆上各点的最大距离为$10cm$，最小距离为$8cm$，则此圆的半径为 (
A.$9cm$
B.$1cm$
C.$9cm$或$1cm$
D.无法确定
C
)A.$9cm$
B.$1cm$
C.$9cm$或$1cm$
D.无法确定
答案
6.C
解析
当点$P$在圆外时，圆的半径$r = \frac{10 - 8}{2} = 1\,cm$；当点$P$在圆内时，圆的半径$r = \frac{10 + 8}{2} = 9\,cm$。故此圆的半径为$9\,cm$或$1\,cm$。
C
C
7.《墨经》有“圆，一中同长也”的记载，它的意思是圆上各点到圆心的距离都等于半径.下列图形中，各边的中点一定在同一个圆上的是 ( )
①矩形；②平行四边形；③菱形；④正方形.
①④
①矩形；②平行四边形；③菱形；④正方形.
答案
7.①④
8.如图24.1.1-12，在$\odot O$中，线段$AB$为其直径，为什么直径$AB$是$\odot O$中最长的弦？
答案
8.解:如答图24.1.1−4,设CD
为⊙O中非直径的任意一
条弦,连接OC,OD,则OC+OD＞CD,而OC,
OD为⊙O的半径，所以直
径大于CD,即直径AB是⊙O中最长的弦.
9.如图24.1.1-13，直线$l$经过$\odot O$的圆心$O$，且与$\odot O$交于$A$，$B$两点，点$C$在$\odot O$上，且$\angle AOC = 30^{\circ}$，点$P$是直线$l$上的一个动点（与圆心$O$不重合），直线$CP$与$\odot O$相交于点$Q$.是否存在点$P$，使得$QP = QO$？若存在，求出相应的$\angle OCP$的大小；若不存在，请简要说明理由.
答案
9.解:存在点P,使得QP=QO,相应的∠OCP分别为40°,100°,20°.
解析
解:存在点$P$,使得$QP = QO$,相应的$\angle OCP$的大小为$20°$、$40°$或$100°$.
情况一：点$P$在$OA$上（不与$O$重合）
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = \angle AOC + \angle OCP = 30° + x$,
$\angle QOP = 30° + x$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC = \angle QOP + \angle AOC$,
$\therefore 180° - 2x = 30° + x + 30°$,
解得$x = 40°$.
情况二：点$P$在$OB$上（不与$O$重合）
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = \angle OCP - \angle AOC = x - 30°$,
$\angle QOP = x - 30°$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC + \angle QOP = 180°$,
$\therefore 180° - 2x + x - 30° = 180°$,
解得$x = -30°$（舍去）.
子情况：点$Q$在优弧$AC$上
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = 30° - x$（$x ＜ 30°$）,
$\angle QOP = 30° - x$.
在$\triangle OQC$中,$\angle QOC = 2x - 180°$.
$\because \angle QOC = \angle AOC - \angle QOP$,
$\therefore 2x - 180° = 30° - (30° - x)$,
解得$x = 20°$.
情况三：点$P$在$OA$延长线上
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = x - 30°$,$\angle QOP = x - 30°$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOP = \angle QOC + \angle AOC$,
$\therefore x - 30° = 180° - 2x + 30°$,
解得$x = 80°$（矛盾，舍去）.
情况四：点$P$在$OB$延长线上
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = 180° - x - 30° = 150° - x$,
$\angle QOP = 150° - x$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC + \angle QOP = 180°$,
$\therefore 180° - 2x + 150° - x = 180°$,
解得$x = 50°$（矛盾，舍去）.
子情况：点$Q$在劣弧$AC$上
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = x - 30°$,$\angle QOP = x - 30°$.
$\angle QOC = 2x - 180°$.
$\because \angle QOC = \angle QOP - \angle AOC$,
$\therefore 2x - 180° = x - 30° - 30°$,
解得$x = 120°$（矛盾，舍去）.
情况五：点$P$与$C$、$Q$构成钝角三角形
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO = \frac{180° - (180° - x)}{2} = \frac{x}{2}$.
$\angle QOP + \angle AOC = \angle OQC$,
$\frac{x}{2} + 30° = 180° - x$,
解得$x = 100°$.
综上，$\angle OCP$的度数为$20°$、$40°$或$100°$.
情况一：点$P$在$OA$上（不与$O$重合）
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = \angle AOC + \angle OCP = 30° + x$,
$\angle QOP = 30° + x$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC = \angle QOP + \angle AOC$,
$\therefore 180° - 2x = 30° + x + 30°$,
解得$x = 40°$.
情况二：点$P$在$OB$上（不与$O$重合）
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = \angle OCP - \angle AOC = x - 30°$,
$\angle QOP = x - 30°$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC + \angle QOP = 180°$,
$\therefore 180° - 2x + x - 30° = 180°$,
解得$x = -30°$（舍去）.
子情况：点$Q$在优弧$AC$上
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = 30° - x$（$x ＜ 30°$）,
$\angle QOP = 30° - x$.
在$\triangle OQC$中,$\angle QOC = 2x - 180°$.
$\because \angle QOC = \angle AOC - \angle QOP$,
$\therefore 2x - 180° = 30° - (30° - x)$,
解得$x = 20°$.
情况三：点$P$在$OA$延长线上
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = x - 30°$,$\angle QOP = x - 30°$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOP = \angle QOC + \angle AOC$,
$\therefore x - 30° = 180° - 2x + 30°$,
解得$x = 80°$（矛盾，舍去）.
情况四：点$P$在$OB$延长线上
设$\angle OCP = x$,则$\angle OQC = x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = 180° - x - 30° = 150° - x$,
$\angle QOP = 150° - x$.
在$\triangle OQC$中,$\angle QOC = 180° - 2x$.
$\because \angle QOC + \angle QOP = 180°$,
$\therefore 180° - 2x + 150° - x = 180°$,
解得$x = 50°$（矛盾，舍去）.
子情况：点$Q$在劣弧$AC$上
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO$.
$\angle QPO = x - 30°$,$\angle QOP = x - 30°$.
$\angle QOC = 2x - 180°$.
$\because \angle QOC = \angle QOP - \angle AOC$,
$\therefore 2x - 180° = x - 30° - 30°$,
解得$x = 120°$（矛盾，舍去）.
情况五：点$P$与$C$、$Q$构成钝角三角形
设$\angle OCP = x$,则$\angle OQC = 180° - x$.
$\because QP = QO$,$\therefore \angle QOP = \angle QPO = \frac{180° - (180° - x)}{2} = \frac{x}{2}$.
$\angle QOP + \angle AOC = \angle OQC$,
$\frac{x}{2} + 30° = 180° - x$,
解得$x = 100°$.
综上，$\angle OCP$的度数为$20°$、$40°$或$100°$.
