《2025年实验班提优训练九年级数学上册人教版》

9. (湖北黄冈中学自主招生)如图,已知$A(12,0),B(8,6)$为平面直角坐标系内两点,以点B为圆心的$\odot B$经过原点O,$BC⊥x$轴于点C,点D为$\odot B$上一动点,E为AD的中点,则线段CE长度的最大值为______.

答案：
5 + $\sqrt{13}$ [解析]如图,取OC的中点M,连接DM,BD,BM,OB.
∵点B的坐标是(8,6),
∴MC = $\frac{1}{2}$ OC = $\frac{1}{2}$ ×8 = 4.
∵点A的坐标是(12,0),
　
∴AC = OA - OC = 12 - 8 = 4,
∴C是AM的中点.
∵E是AD的中点,
∴CE是△ADM的中位线,
∴CE = $\frac{1}{2}$ DM.
∵DM≤BD + BM,
∴CE≤ $\frac{1}{2}$ (BD + BM).
∵BC⊥OA,
∴∠BCM = 90°,
∴MB = $\sqrt{MC^{2}+BC^{2}}$ = $\sqrt{4^{2}+6^{2}}$ = 2$\sqrt{13}$,OB = $\sqrt{BC^{2}+OC^{2}}$ = $\sqrt{8^{2}+6^{2}}$ = 10,
∴BD = OB = 10,
∴CE≤ $\frac{1}{2}$ ×(10 + 2$\sqrt{13}$) = 5 + $\sqrt{13}$,
∴CE的最大值为5 + $\sqrt{13}$.
　一题多解　如图,连接AB,BD,取AB的中点J,连接EJ,CJ.
　
∵AB = $\sqrt{BC^{2}+AC^{2}}$ = 2$\sqrt{13}$,
∴CJ = $\frac{1}{2}$ AB = $\sqrt{13}$.
∵AJ = JB,AE = ED,
∴EJ = $\frac{1}{2}$ BD = $\frac{1}{2}$ OB = $\frac{1}{2}$ $\sqrt{OC^{2}+BC^{2}}$ = 5,
∴CE≤JE + CJ = 5 + $\sqrt{13}$,
∴CE的最大值为5 + $\sqrt{13}$.

10. 如图,在平面直角坐标系中,A,B,C是$\odot M$上的三个点,$A(0,4),B(4,4),C(6,2)$.
(1)圆心M的坐标为______;
(2)判断点$D(4,-3)与\odot M$的位置关系.

答案：

(1)(2,0)
(2)如图,
　
　圆的半径AM = $\sqrt{2^{2}+4^{2}}$ = 2$\sqrt{5}$,MD = $\sqrt{(4 - 2)^{2}+3^{2}}$ = $\sqrt{13}$＜2$\sqrt{5}$,所以点D在⊙M内.

11. (2024·江苏南京建邺区期中)如图,在$\triangle ABC$中,$AB= AC,\odot O是\triangle ABC$的外接圆,过点O作AC的垂线,垂足为D,分别交CB的延长线,$\overset{\frown }{AC}$于点E,F,$AF$,BC的延长线交于点G.
(1)求证:$AC= CG$;
(2)若$EB= CG$,求$∠BAC$的度数.

答案：

(1)如图,连接CF.
∵过点O作AC的垂线,垂足为D,
∴ $\overarc{AF}$ = $\overarc{CF}$,
∴∠FAC = ∠ACF.
∵AB = AC,
∴∠ABC = ∠ACB.
∵∠ABC + ∠AFC = 180°,∠GFC + ∠AFC = 180°,
∴∠GFC = ∠ABC,
∴∠GFC = ∠ACB.
∵∠G = ∠ACB - ∠FAC,∠ACF = ∠GFC - ∠FAC,
∴∠ACF = ∠G,
∴∠CAF = ∠G,
∴AC = CG.
　
(2)如图,连接AE,过点A作AH⊥BC于点H.
∵AB = AC,
∴BH = CH, $\overarc{AB}$ = $\overarc{AC}$,
∴AH过点O.
∵BE = CG,
∴EH = GH,
∴AE = AG,AB = BE = AC = CG,
∴∠AEB = ∠BAE = ∠G = ∠CAG.
∵EF⊥AC,AD = CD,
∴AE = EC,
∴∠AED = ∠CED.设∠AED = ∠CED = α,
∴∠ACB = ∠ABC = ∠AEB + ∠BAE = 4α.
∵∠CDE = 90°,
∴α + 4α = 90°,
∴α = 18°,
∴∠ABC = ∠ACB = 72°,
∴∠BAC = 180° - ∠ABC - ∠ACB = 36°.

12. (2025·湖北武汉江夏区期中)如图,$\triangle ABC是\odot O$的内接三角形,点D是$\overset{\frown }{AB}$的中点,连接AD,BD,CD.
(1)如图(1),若$∠ACD= 30^{\circ }$,求$∠ADB$的度数;
(2)如图(2),若$AB= AC= 10,AD= \frac {5\sqrt {5}}{2}$,求BC的长.

答案：

(1)
∵点D是 $\overarc{AB}$ 的中点,
∴ $\overarc{BD}$ = $\overarc{AD}$,
∴∠ACD = ∠BCD = 30°,
∴∠ACB = 60°,
∴∠ADB = 180° - ∠ACB = 120°.
(2)如图,连接AO并延长交BC于点E,连接OD交AB于点H.
∵AB = AC,
∴ $\overarc{AB}$ = $\overarc{AC}$,
∴AE⊥BC,
∴BE = CE.
∵点D是 $\overarc{AB}$ 的中点,
　
∴OD⊥AB,AH = $\frac{1}{2}$ AB = 5,
∴DH = $\sqrt{AD^{2}-AH^{2}}$ = $\sqrt{(\frac{5\sqrt{5}}{2})^{2}-5^{2}}$ = $\frac{5}{2}$.
　利用勾股定理求三角形直角边
　在Rt△AOH中,OA² = AH² + OH²,
∴OA² = 5² + (OA - $\frac{5}{2}$)²,解得OA = OB = $\frac{25}{4}$.
∵OB² - OE² = BE² = AB² - AE²,
∴($\frac{25}{4}$)² - OE² = 10² - ($\frac{25}{4}$ + OE)²,解得OE = $\frac{7}{4}$,
∴BE = $\sqrt{OB^{2}-OE^{2}}$ = $\sqrt{(\frac{25}{4})^{2}-(\frac{7}{4})^{2}}$ = 6,
∴BC = 2BE = 12.

13. (2024·广州中考)如图,$\odot O$中,弦AB的长为$4\sqrt {3}$,点C在$\odot O$上,$OC⊥AB,∠ABC= 30^{\circ }.\odot O$所在的平面内有一点P,若$OP= 5$,则点P与$\odot O$的位置关系是(C).

A. 点P在$\odot O$上
B. 点P在$\odot O$内
C. 点P在$\odot O$外
D. 无法确定

答案： C [解析]设AB与OC交于点D.
∵弦AB的长为4$\sqrt{3}$,OC⊥AB,
∴AD = BD = $\frac{1}{2}$ AB = 2$\sqrt{3}$,∠ADO = 90°.
∵∠ABC = 30°,
∴∠AOD = 2∠ABC = 60°,
∴∠A = 90° - 60° = 30°,
∴OA = 2OD.
　设OD = x,则OA = 2x,
　在Rt△AOD中,OD² + AD² = OA²,
　即x² + (2$\sqrt{3}$)² = (2x)²,
　解得x = 2(负值已舍去),
∴OA = 2x = 4.
∵OP = 5,
∴OP＞OA,
∴点P在⊙O外.故选C.