1. [2026 泰州靖江期中]如图,点A,B,C是$\odot O$上不重合的三点,则下列结论一定正确的是 ()

A.$∠ AOB = ∠ A + ∠ B$
B.$∠ AOB = 2(∠ A + ∠ B)$
C.$∠ AOB = 90° - (∠ A + ∠ B)$
D.$∠ AOB = 180° - 2(∠ A + ∠ B)$

1 B 如图,连接CO.
∵ OA = OB = OC,
∴ ∠A = ∠OCA,∠B = ∠OCB,
∴ ∠ACB = ∠OCA + ∠OCB = ∠A + ∠B,
∴ ∠AOB = 2∠ACB = 2(∠A + ∠B).

2. [2026南京弘光中学月考]如图，AB为$\odot O$的直径，CD是$\odot O$的弦，AB，CD的延长线交于点E. 已知$AB=2DE$，$∠ E=16°$，则$∠ AOC$的度数是°.

2 48 连接OD.
∵ AB 为$\odot O$的直径,
∴ AB = 2OD.
∵ AB = 2DE,
∴ DE = DO,
∴ ∠DOE = ∠E = 16°,
∴ ∠CDO = ∠E + ∠DOE = 32°.
∵ OC = OD,
∴ ∠C = ∠CDO = 32°,
∴ ∠AOC = ∠C + ∠E = 32° + 16° = 48°.

3. [2025 南通通州育才中学月考] 如图,半径为5的$\odot A$与y轴交于点$B(0,2),C(0,10)$,则点A的横坐标为 ()


A.$-3$
B.$3$
C.$4$
D.$6$

3 B 如图,过点A作$AD ⊥ BC$于点D,连接AB.
∵ 半径为5的$\odot A$与y轴交于点$B(0,2),C(0,10)$,
∴ AB = 5,BC = 10 - 2 = 8.
∵ $AD ⊥ BC$,AD 过圆心A,
∴ CD = BD = 4. 在Rt△ADB中,由勾股定理,得$AD = \sqrt{AB^2 - BD^2} = 3$,
∴ 点A的横坐标是3.

4. [2025无锡江阴期中]把球放在长方体纸盒内,球的一部分露出盒外,其截面如图所示,已知$EF=CD=8\ \mathrm{cm}$,则球的半径是 ()


A.4 cm
B.5 cm
C.6 cm
D.8 cm

4 B 设球心为O,如图,过点O作$ON ⊥ AD$于点N,延长NO交CB于点M,连接OF.
∵ 四边形ABCD是矩形,
∴ ∠C = ∠D = 90°,
∴ 四边形CDNM是矩形,
∴ MN = CD = 8 cm. 设OF = x cm,则OM = OF = x cm,
∴ ON = MN - OM = (8 - x) cm.
∵ EF = 8 cm,
∴ NF = EN = 4 cm. 在Rt△ONF中,$ON^2 + NF^2 = OF^2$,即$(8 - x)^2 + 4^2 = x^2$,解得x = 5,则球的半径是5 cm.

5. [2026宿迁宿豫区月考] 已知$\odot O$的半径为13，弦$AB// CD$，$AB=24$，$CD=10$，则两弦之间的距离为。

5 7 或 17 如图1,当弦AB和CD在圆心异侧时,过点O作$OF ⊥ CD$于F,延长FO交AB于E,连接OA,OC.
∵ AB//CD,$OF ⊥ CD$,
∴ $OE ⊥ AB$.
∵ $OF ⊥ CD,CD = 10$,
∴ $CF = \frac{1}{2}CD = \frac{1}{2} × 10 = 5$,$∠ OFC = 90°$,
∴ $OF = \sqrt{OC^2 - CF^2} = \sqrt{13^2 - 5^2} = 12$. 同理可得OE = 5,
∴ EF = OF + OE = 12 + 5 = 17. 如图2,当弦AB和CD在圆心同侧时,过点O作$OF ⊥ CD$于F,交AB于E,连接OA,OC. 同理可得OE = 5,OF = 12,
∴ EF = OF - OE = 12 - 5 = 7. 综上可知,两弦之间的距离为7或17.

6. 新情境 如图，C是AE的中点，在AE同侧分别以AC,CE为直径作半圆B和半圆D，直线l//AE，与两个半圆依次相交于F,M,N,G不同的四点.若AE=10,FG=x,MN=y,则y与x之间的函数表达式为.

6 y = 10 - x 解题思路:过点B作$BQ ⊥ FM$于点Q,过点D作$DH ⊥ NG$于点H,连接BF,DG. 根据垂径定理,得FQ = MQ,NH = GH,再证明四边形BDHQ为矩形,则QH = BD = 5,
∴ QM + NH = 5 - y,接着证明FQ = GH,得到FQ = MQ = NH = GH. 利用等量代换得到FG = 2(QM + NH) + MN,即x = 2(5 - y) + y = 10 - y,进而得解.
如图,过点B作$BQ ⊥ FM$于点Q,过点D作$DH ⊥ NG$于点H,连接BF,DG,则FQ = MQ,NH = GH.
∵ l//AE,
∴ BQ = DH(平行线间的距离处处相等),$BQ ⊥ AE$,$DH ⊥ AE$,
∴ 四边形BDHQ为矩形,
∴ $QH = BD = \frac{1}{2}AE = 5$,即QM + MN + NH = 5,
∴ QM + NH = 5 - y.
∵ $FQ = \sqrt{BF^2 - BQ^2}$,$GH = \sqrt{DG^2 - DH^2}$,而BF = DG,
∴ FQ = GH,
∴ FQ = MQ = NH = GH.
∵ FG = FM + MN + NG = 2QM + MN + 2NH = 2(QM + NH) + MN,
∴ x = 2(5 - y) + y = 10 - y,
∴ y = 10 - x.