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2.5.2 圆与圆的位置关系

发布时间:2024-12-24 06:09

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【基础过关系列】2022-2023 学年高二数学上学期同步知识点剖析精品讲义 (人教 A 版 2019)
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选择性必修第一册同步巩固,难度 2 颗星!

基础知识

圆与圆的位置关系

1 在同一个平面内,两个不等的圆的位置关系共有五种:外离、外切、相交、内切、内含.
圆心距:两圆圆心的距离叫做圆心距.

  2 设两圆 &#x2299;O1" role="presentation">⊙O1 与圆 &#x2299;O2" role="presentation">⊙O2 的圆心距为 O1O2=d" role="presentation">O1O2=d,半径分别为 r" role="presentation">r,R(r&lt;R)" role="presentation">R(r<R),则有

注 当两圆相切时分为两种情况:外切和内切.
 

判断两个圆位置关系的方法

1 几何法
通过判断圆心距与两个圆半径之间的关系从而确定两个圆的位置关系.

  2 代数法
设由两圆的方程组成的方程组为 {x2+y2+D1x+E1y+F1=0x2+y2+D2x+E2y+F2=0" role="presentation">{x2+y2+D1x+E1y+F1=0x2+y2+D2x+E2y+F2=0,
由此方程组得:有两组不同的实数解则两圆相交;有两组相同的实数解则两圆相切;无实数解则两圆相离.

【例】判断圆 &#x2299;O1:x2+y2=1" role="presentation">⊙O1:x2+y2=1 与 &#x2299;O2:x2+y2&#x2212;2x&#x2212;4y+1=0" role="presentation">⊙O2:x2+y2−2x−4y+1=0 的位置关系.
解析 几何法
圆 x2+y2=1" role="presentation">x2+y2=1 圆心 O1(0,0)" role="presentation">O1(0,0),半径 r=1" role="presentation">r=1,
&#x2299;O2:x2+y2&#x2212;2x&#x2212;4y+1=0&#x27F9;(x&#x2212;1)2+(y&#x2212;2)2=4" role="presentation">⊙O2:x2+y2−2x−4y+1=0⟹(x−1)2+(y−2)2=4 圆心 O2(1,2)" role="presentation">O2(1,2),半径 r=2" role="presentation">r=2,
则圆心距 O1O2=(1&#x2212;0)2+(2&#x2212;0)2=5" role="presentation">O1O2=(1−0)2+(2−0)2=5,R&#x2212;r=3" role="presentation">R−r=3,R&#x2212;r=1" role="presentation">R−r=1,
则 R&#x2212;r&lt;O1O2&lt;R&#x2212;r" role="presentation">R−r<O1O2<R−r,故两个圆相交.
代数法
联立方程 {x2+y2=1x2+y2&#x2212;2x&#x2212;4y+1=0" role="presentation">{x2+y2=1x2+y2−2x−4y+1=0 得 5y2&#x2212;4y=0" role="presentation">5y2−4y=0,显然方程组有两个实数解,
故两个圆相交.
 

两圆的公共弦方程

(1) 两圆 C1&#xFF1A;x2+y2+D1x+E1y+F1=0" role="presentation">C1:x2+y2+D1x+E1y+F1=0, C2&#xFF1A;x2+y2+D2x+E2y+F2=0" role="presentation">C2:x2+y2+D2x+E2y+F2=0 相交,交点为 A" role="presentation">A,B" role="presentation">B,
则线段 AB" role="presentation">AB 为两圆的公共弦;

(2) 若两圆 C1&#xFF1A;x2+y2+D1x+E1y+F1=0" role="presentation">C1:x2+y2+D1x+E1y+F1=0, C2&#xFF1A;x2+y2+D2x+E2y+F2=0" role="presentation">C2:x2+y2+D2x+E2y+F2=0 相交于 A" role="presentation">A,B" role="presentation">B 两点,
则直线 AB" role="presentation">AB 的方程为 (D1&#x2212;D2)x+(E1&#x2212;E2)y+F1&#x2212;F2=0" role="presentation">(D1−D2)x+(E1−E2)y+F1−F2=0.
证明 设点 A(x1,y1)" role="presentation">A(x1,y1),B(x2,y2)" role="presentation">B(x2,y2),
因为点 A" role="presentation">A 在两圆上,
所以 x12+y12+D1x1+E1y1+F1=0" role="presentation">x12+y12+D1x1+E1y1+F1=0,且 x12+y12+D2x1+E2y1+F2=0" role="presentation">x12+y12+D2x1+E2y1+F2=0,
两等式相减得 (D1&#x2212;D2)x1+(E1&#x2212;E2)y1+F1&#x2212;F2=0(1)" role="presentation">(D1−D2)x1+(E1−E2)y1+F1−F2=0(1),
同理可得 (D1&#x2212;D2)x2+(E1&#x2212;E2)y2+F1&#x2212;F2=0(2)" role="presentation">(D1−D2)x2+(E1−E2)y2+F1−F2=0(2),
由 (1),(2) 可知点 A" role="presentation">A,B" role="presentation">B 均在直线 (D1&#x2212;D2)x+(E1&#x2212;E2)y+F1&#x2212;F2=0" role="presentation">(D1−D2)x+(E1−E2)y+F1−F2=0,
即直线 AB" role="presentation">AB 的方程为 (D1&#x2212;D2)x+(E1&#x2212;E2)y+F1&#x2212;F2=0" role="presentation">(D1−D2)x+(E1−E2)y+F1−F2=0.
 

【例】若圆 x2+y2=1" role="presentation">x2+y2=1 与 x2+y2+2x&#x2212;2y=0" role="presentation">x2+y2+2x−2y=0 相交于 A" role="presentation">A,B" role="presentation">B 两点,求直线 AB" role="presentation">AB 的方程.

解析 两圆相减: x2+y2+2x&#x2212;2y&#x2212;(x2+y2)=0&#x2212;1" role="presentation">x2+y2+2x−2y−(x2+y2)=0−1,化简得 2x&#x2212;2y+1=0" role="presentation">2x−2y+1=0,
即直线 AB" role="presentation">AB 的方程为 2x&#x2212;2y+1=0" role="presentation">2x−2y+1=0.
 

基本方法

【题型1】两圆的位置关系

【典题 1】 已知圆 C1:x2+y2&#x2212;2ax&#x2212;2y+a2&#x2212;15=0" role="presentation">C1:x2+y2−2ax−2y+a2−15=0,圆 C2:x2+y2&#x2212;4ax&#x2212;2y+4a2=0(a&gt;0)" role="presentation">C2:x2+y2−4ax−2y+4a2=0(a>0).试求 a" role="presentation">a 为何值时两圆 C1" role="presentation">C1,C2" role="presentation">C2 的位置关系为:
  (1) 相切; " role="presentation"> (2) 相交;" role="presentation"> (3) 外离; " role="presentation"> (4) 内含.
解析 圆 C1" role="presentation">C1,C2" role="presentation">C2 的方程,经配方后可得: C1:(x&#x2212;a)2+(y&#x2212;1)2=16" role="presentation">C1:(x−a)2+(y−1)2=16,C2:(x&#x2212;2a)2+(y&#x2212;1)2=1" role="presentation">C2:(x−2a)2+(y−1)2=1,
&#x2234;" role="presentation">∴ 圆心 C1(a,1)" role="presentation">C1(a,1),C2(2a,1)" role="presentation">C2(2a,1),半径 r1=4" role="presentation">r1=4,r2=1" role="presentation">r2=1.
&#x2234;|C1C2|=(a&#x2212;2a)2+(1&#x2212;1)2=a" role="presentation">∴|C1C2|=(a−2a)2+(1−1)2=a.
(1) 当 |C1C2|=r1+r2=5" role="presentation">|C1C2|=r1+r2=5,即 a=5" role="presentation">a=5 时,两圆外切,
当 |C1C2|=r1&#x2212;r2=3" role="presentation">|C1C2|=r1−r2=3,即 a=3" role="presentation">a=3 时,两圆内切.
(2) 当 3&lt;|C1C2|&lt;5" role="presentation">3<|C1C2|<5,即 3&lt;a&lt;5" role="presentation">3<a<5 时,两圆相交.
(3) 当 |C1C2|&gt;5" role="presentation">|C1C2|>5,即 a&gt;5" role="presentation">a>5 时,两圆外离.
(4) 当 |C1C2|&lt;3" role="presentation">|C1C2|<3,即 a&lt;3" role="presentation">a<3 时,两圆内含.
 

巩固练习

1. 已知圆 A" role="presentation">A,圆 B" role="presentation">B 相切,圆心距为 10cm" role="presentation">10cm,其中圆 A" role="presentation">A 的半径为 4cm" role="presentation">4cm,则圆 B" role="presentation">B 的半径为 &#x005F;" role="presentation">_ .
 

2. 圆 (x+2)2+y2=4" role="presentation">(x+2)2+y2=4 与圆 (x&#x2212;2)2+(y&#x2212;1)2=9" role="presentation">(x−2)2+(y−1)2=9 的位置关系为 &#x005F;" role="presentation">_.
 

3. 两圆 x2+y2=1" role="presentation">x2+y2=1 和 (x&#x2212;1)2+(y&#x2212;a)2=4" role="presentation">(x−1)2+(y−a)2=4 相切,则实数 a" role="presentation">a 的值 &#x005F;" role="presentation">_.
 

4. 已知点 O(0,0)" role="presentation">O(0,0),M(1,0)" role="presentation">M(1,0),且圆 C&#xFF1A;(x&#x2212;5)2+(y&#x2212;4)2=r2(r&gt;0)" role="presentation">C:(x−5)2+(y−4)2=r2(r>0) 上至少存在一点 P" role="presentation">P,使得 |PO|=2|PM|" role="presentation">|PO|=2|PM|,则 r" role="presentation">r 的最小值是 &#x005F;" role="presentation">_.
 

参考答案

答案 6cm" role="presentation">6cm 或 14cm" role="presentation">14cm
解析 设圆 B" role="presentation">B 的半径为 rcm" role="presentation">rcm,
因为圆 A" role="presentation">A、圆 B" role="presentation">B 相切,圆心距为 10cm" role="presentation">10cm,
①当圆 A" role="presentation">A 与圆 B" role="presentation">B 外切时,则 r+4=10" role="presentation">r+4=10,解得 r=6" role="presentation">r=6;
②当圆 A" role="presentation">A 与圆 B" role="presentation">B 内切时,则 |r&#x2212;4|=10" role="presentation">|r−4|=10,解得 r=14" role="presentation">r=14.
故圆 B 半径为 6cm" role="presentation">6cm 或 14cm" role="presentation">14cm. 答案 相交
解析 圆 (x+2)2+y2=4" role="presentation">(x+2)2+y2=4 的圆心 C1(&#x2212;2&#xFF0C;0)" role="presentation">C1(−2,0),半径 r=2" role="presentation">r=2.
圆 (x&#x2212;2)2+(y&#x2212;1)2=9" role="presentation">(x−2)2+(y−1)2=9 的圆心 C2(2&#xFF0C;1)" role="presentation">C2(2,1),半径 R=3" role="presentation">R=3,
两圆的圆心距离 d=(&#x2212;2&#x2212;2)2+(0&#x2212;1)2=17" role="presentation">d=(−2−2)2+(0−1)2=17,
R+r=5" role="presentation">R+r=5,R&#x2212;r=1" role="presentation">R−r=1,R+r&gt;d&gt;R&#x2212;r" role="presentation">R+r>d>R−r,
所以两圆相交. 答案 a=&#x00B1;22" role="presentation">a=±22 或 a=0" role="presentation">a=0
解析 两圆圆心距为 a2+1" role="presentation">a2+1,因为两圆相切,
所以 a2+1=2+1" role="presentation">a2+1=2+1 或 a2+1=2&#x2212;1" role="presentation">a2+1=2−1,
即 a2+1=3" role="presentation">a2+1=3 或 a2+1=1" role="presentation">a2+1=1.所以 a=&#x00B1;22" role="presentation">a=±22 或 a=0" role="presentation">a=0. 答案 5&#x2212;2" role="presentation">5−2
解析 设 P(x,y)" role="presentation">P(x,y),
&#x2235;|PO|=2|PM|" role="presentation">∵|PO|=2|PM|,
&#x2234;x2+y2=2(x&#x2212;1)2+2y2" role="presentation">∴x2+y2=2(x−1)2+2y2,即 (x&#x2212;2)2+y2=2" role="presentation">(x−2)2+y2=2,
圆心距 =(5&#x2212;2)2+(4&#x2212;0)2=r+2" role="presentation">=(5−2)2+(4−0)2=r+2,
&#x2234;r" role="presentation">∴r 的最小值是 5&#x2212;2" role="presentation">5−2.
 

【题型2】两圆公共弦问题

【典题 1】 已知圆 C1:x2+y2+6x&#x2212;4=0" role="presentation">C1:x2+y2+6x−4=0 和圆 C2:x2+y2+6y&#x2212;28=0" role="presentation">C2:x2+y2+6y−28=0.
  (1) 求两圆公共弦所在直线的方程;
  (2) 求经过两圆交点且圆心在直线 x&#x2212;y&#x2212;4=0" role="presentation">x−y−4=0 上的圆的方程.
解析 (1) 设两圆交点为 A(x1,y1)" role="presentation">A(x1,y1),B(x2,y2)" role="presentation">B(x2,y2),
则 A" role="presentation">A,B" role="presentation">B 两点坐标是方程组 {x2+y2+6x&#x2212;4=0(1)x2+y2+6y&#x2212;28=0(2)" role="presentation">{x2+y2+6x−4=0(1)x2+y2+6y−28=0(2) 的解.
①&#xFF0D;" role="presentation">-②得 x&#x2212;y+4=0" role="presentation">x−y+4=0.
&#x2235;A,B" role="presentation">∵A,B 两点坐标都满足此方程,
&#x2234;x&#x2212;y+4=0" role="presentation">∴x−y+4=0 即为两圆公共弦所在直线的方程.
(2) 方法一:解方程组 {x2+y2+6x&#x2212;4=0x2+y2+6y&#x2212;28=0" role="presentation">{x2+y2+6x−4=0x2+y2+6y−28=0 两圆的交点 A(&#x2212;1,3)" role="presentation">A(−1,3),B(&#x2212;6,&#x2212;2)" role="presentation">B(−6,−2).
设所求圆的圆心为 (a,b)" role="presentation">(a,b),因圆心在直线 x&#x2212;y&#x2212;4=0" role="presentation">x−y−4=0 上,
故 b=a&#xFF0D;4" role="presentation">b=a-4.
则 (a+1)2+(a&#x2212;4&#x2212;3)2=(a+6)2+(a&#x2212;4+2)2" role="presentation">(a+1)2+(a−4−3)2=(a+6)2+(a−4+2)2,解得 a=12" role="presentation">a=12,
故圆心为 (12,&#x2212;72)" role="presentation">(12,−72),半径为 892" role="presentation">892.
故圆的方程为 (x&#x2212;12)2+(y+72)2=892" role="presentation">(x−12)2+(y+72)2=892,
即 x2+y2&#x2212;x+7y&#x2212;32=0" role="presentation">x2+y2−x+7y−32=0.
方法二:设所求圆的方程为 x2+y2+6x&#x2212;4+&#x3BB;(x2+y2+6y&#x2212;28)=0(&#x3BB;&#x2260;&#x2212;1)" role="presentation">x2+y2+6x−4+λ(x2+y2+6y−28)=0(λ≠−1),
其圆心为 (&#x2212;31+&#x03BB;,&#x2212;3&#x03BB;1+&#x03BB;)" role="presentation">(−31+λ,−3λ1+λ),代入 x&#x2212;y&#x2212;4=0" role="presentation">x−y−4=0,解得 &#x3BB;=&#x2212;7" role="presentation">λ=−7.
故所求圆的方程为 x2+y2&#x2212;x+7y&#x2212;32=0" role="presentation">x2+y2−x+7y−32=0.
点拨 过两圆 C1&#xFF1A;x2+y2+D1x+E1y+F1=0" role="presentation">C1:x2+y2+D1x+E1y+F1=0, C2&#xFF1A;x2+y2+D2x+E2y+F2=0" role="presentation">C2:x2+y2+D2x+E2y+F2=0 交点的圆系方程为 x2+y2+D1x+E1y+F1+&#x3BB;(x2+y2+D2x+E2y+F2)=0" role="presentation">x2+y2+D1x+E1y+F1+λ(x2+y2+D2x+E2y+F2)=0
(&#x3BB;&#x2260;&#x2212;1" role="presentation">λ≠−1, 此圆系不含 C2&#xFF1A;x2+y2+D2x+E2y+F2=0" role="presentation">C2:x2+y2+D2x+E2y+F2=0)
特别地,当 &#x3BB;=&#x2212;1" role="presentation">λ=−1 时,上述方程为一次方程.
两圆相交时,表示公共弦方程;两圆相切时,表示公切线方程.
 

巩固练习

1.(多选) 已知圆 C1&#xFF1A;x2+y2=1" role="presentation">C1:x2+y2=1,圆 C2&#xFF1A;(x&#x2212;3)2+(y+4)2=r2(r&gt;0)" role="presentation">C2:(x−3)2+(y+4)2=r2(r>0),则 (  )
 A.若圆 C1" role="presentation">C1 与圆 C2" role="presentation">C2 无公共点,则 0&lt;r&lt;4" role="presentation">0<r<4
 B.当 r=5" role="presentation">r=5 时,两圆公共弦长所在直线方程为 6x&#x2212;8y&#x2212;1=0" role="presentation">6x−8y−1=0
 C.当 r=2" role="presentation">r=2 时,P" role="presentation">P、Q" role="presentation">Q 分别是圆 C1" role="presentation">C1 与圆 C2" role="presentation">C2 上的点,则 |PQ|" role="presentation">|PQ| 的取值范围为 [2,8]" role="presentation">[2,8]
 D.当 0&lt;r&lt;4" role="presentation">0<r<4 时,过直线 6x&#x2212;8y+r2&#xFF0D;26=0" role="presentation">6x−8y+r2-26=0 上任意一点分别作圆 C1" role="presentation">C1、圆 C2" role="presentation">C2 切线,则切线长相等
 

2. 圆 x2+y2&#x2212;2x&#x2212;5=0" role="presentation">x2+y2−2x−5=0 和圆 x2+y2+2x&#x2212;4y&#x2212;4=0" role="presentation">x2+y2+2x−4y−4=0 的交点为 A" role="presentation">A,B" role="presentation">B,则线段 AB" role="presentation">AB 的垂直平分线的方程为 &#x005F;" role="presentation">_.
 

3. 如图所示,在圆 O" role="presentation">O 上任取 C" role="presentation">C 点为圆心,作一圆 C" role="presentation">C 与圆 O" role="presentation">O 的直径 AB" role="presentation">AB 相切于 D" role="presentation">D,圆 C" role="presentation">C 与圆 O" role="presentation">O 交于 E" role="presentation">E,F" role="presentation">F,且 EF" role="presentation">EF 与 CD" role="presentation">CD 相交于 H" role="presentation">H.

求证:EF" role="presentation">EF 平分 CD" role="presentation">CD.
 

参考答案

答案 BCD" role="presentation">BCD
解析 圆 C1&#xFF1A;x2+y2=1" role="presentation">C1:x2+y2=1,其圆心为 (0,0)" role="presentation">(0,0),半径为 1" role="presentation">1,
圆 C2&#xFF1A;(x&#xFF0D;3)2+(y+4)2=r2(r&gt;0)" role="presentation">C2:(x-3)2+(y+4)2=r2(r>0),其圆心为 (3,4)" role="presentation">(3,4),半径为 r" role="presentation">r.
对于 A" role="presentation">A,两圆的圆心距 |C1C2|=9+16=5" role="presentation">|C1C2|=9+16=5,
若圆 C1" role="presentation">C1 与圆 C2" role="presentation">C2 无公共点,即两圆内含或外离,
必有 r+1&lt;5" role="presentation">r+1<5 或 r&#x2212;1&gt;5" role="presentation">r−1>5,即 0&lt;r&lt;4" role="presentation">0<r<4 或 r&gt;6" role="presentation">r>6,故 A" role="presentation">A 错误;
对于 B" role="presentation">B,当 r=5" role="presentation">r=5 时,圆 C2&#xFF1A;(x&#x2212;3)2+(y+4)2=25" role="presentation">C2:(x−3)2+(y+4)2=25,
联立圆 C1" role="presentation">C1 和圆 C2" role="presentation">C2 的方程可得 6x&#x2212;8y&#x2212;1=0" role="presentation">6x−8y−1=0,
即两圆公共弦长所在直线方程为 6x&#x2212;8y&#x2212;1=0" role="presentation">6x−8y−1=0,B" role="presentation">B 正确;
对于 C" role="presentation">C,当 r=2" role="presentation">r=2 时,圆 C2&#xFF1A;&#xFF08;x&#x2212;3&#xFF09;2+(y+4)2=r2" role="presentation">C2:(x−3)2+(y+4)2=r2,
P" role="presentation">P、Q" role="presentation">Q 分别是圆 C1" role="presentation">C1 与圆 C2" role="presentation">C2 上的点,
则 |PQ|" role="presentation">|PQ| 的最小值为 |C1C2|&#x2212;2&#x2212;1=2" role="presentation">|C1C2|−2−1=2,最大值为 |C1C2|+2+1=8" role="presentation">|C1C2|+2+1=8,
即 |PQ|" role="presentation">|PQ| 的取值范围为 [2,8]" role="presentation">[2,8],故 C" role="presentation">C 正确;
对于 D" role="presentation">D,当 0&lt;r&lt;4" role="presentation">0<r<4 时,两圆外离,
过 P(x0,y0)" role="presentation">P(x0,y0) 引两圆 C1" role="presentation">C1,C2" role="presentation">C2 的切线分别为 PA" role="presentation">PA,PB" role="presentation">PB,
由 |PA|=|PB|" role="presentation">|PA|=|PB|,得 |PC1|2&#x2212;1=|PC2|2&#x2212;r2" role="presentation">|PC1|2−1=|PC2|2−r2,
即 x02+y02&#x2212;1=(x0&#x2212;3)2+(y0+4)2&#x2212;r2" role="presentation">x02+y02−1=(x0−3)2+(y0+4)2−r2,
整理得: 6x0&#x2212;8y0+r2&#x2212;26=0&#xFF0C;" role="presentation">6x0−8y0+r2−26=0,,
可得过直线 6x&#xFF0D;8y+r2&#xFF0D;26=0" role="presentation">6x-8y+r2-26=0 上任意一点分别作圆 C1" role="presentation">C1、圆 C2" role="presentation">C2 切线,则切线长相等,
故 D" role="presentation">D 正确.
故选:BCD" role="presentation">BCD.

答案 x+y&#x2212;1=0" role="presentation">x+y−1=0
解析 &#x2235;" role="presentation">∵ 圆 x2+y2&#x2212;2x&#x2212;5=0" role="presentation">x2+y2−2x−5=0 的圆心 M(1,0)" role="presentation">M(1,0),
圆 x2+y2+2x&#x2212;4y&#x2212;4=0" role="presentation">x2+y2+2x−4y−4=0 的圆心 N(&#x2212;1,2)" role="presentation">N(−1,2),
又 &#x2235;" role="presentation">∵ 圆 x2+y2&#x2212;2x&#x2212;5=0" role="presentation">x2+y2−2x−5=0 和圆 x2+y2+2x&#x2212;4y&#x2212;4=0" role="presentation">x2+y2+2x−4y−4=0 的交点为 A" role="presentation">A,B" role="presentation">B,
&#x2234;" role="presentation">∴ 线段 AB" role="presentation">AB 的垂直平分线是 MN" role="presentation">MN 所在的直线,
&#x2234;MN" role="presentation">∴MN 的方程为: y&#x2212;02&#x2212;0=x&#x2212;1&#x2212;1&#x2212;1" role="presentation">y−02−0=x−1−1−1,即 x+y&#x2212;1=0" role="presentation">x+y−1=0.

证明 以 AB" role="presentation">AB 所在直线为 x" role="presentation">x 轴,O" role="presentation">O 为坐标原点建立平面直角坐标系.

如图,设 |AB|=2r" role="presentation">|AB|=2r,D(a,0)" role="presentation">D(a,0),
则 |CD|=r2&#x2212;a2" role="presentation">|CD|=r2−a2, &#x2234;C(a,r2&#x2212;a2)" role="presentation">∴C(a,r2−a2).
&#x2234;" role="presentation">∴ 圆 O:x2+y2=r2" role="presentation">O:x2+y2=r2,圆 C:(x&#x2212;a)2+(y&#x2212;r2&#x2212;a2)2=r2&#x2212;a2" role="presentation">C:(x−a)2+(y−r2−a2)2=r2−a2.
两方程作差得直线 EF" role="presentation">EF 的方程为 2ax+2r2&#x2212;a2y=r2+a2" role="presentation">2ax+2r2−a2y=r2+a2.
令 x=a" role="presentation">x=a,得 y=12r2&#x2212;a2" role="presentation">y=12r2−a2,
&#x2234;H(a,12r2&#x2212;a2)" role="presentation">∴H(a,12r2−a2),即 H" role="presentation">H 为 CD" role="presentation">CD 的中点.
&#x2234;EF" role="presentation">∴EF 平分 CD" role="presentation">CD.
 

分层练习

【A组---基础题】

1. 圆 O1:x2+y2=2" role="presentation">O1:x2+y2=2 和圆 O2:x2+y2+4y+3=0" role="presentation">O2:x2+y2+4y+3=0 的位置关系是 (  )
 A.相离 " role="presentation"> B.外切 " role="presentation"> C.内切 " role="presentation">D.相交
 

2. 设 r&gt;0" role="presentation">r>0,两圆 (x&#x2212;1)2+(y+3)2=r2" role="presentation">(x−1)2+(y+3)2=r2 与 x2+y2=16" role="presentation">x2+y2=16 的位置关系不可能是 (  )
  A.相切 " role="presentation"> B.相交 " role="presentation"> C.内切和内含 " role="presentation"> D.外切和外离

  3.(多选) 点 P" role="presentation">P 在圆 C1&#xFF1A;x2+y2=1" role="presentation">C1:x2+y2=1 上,点 Q" role="presentation">Q 在圆 C2&#xFF1A;x2+y2&#xFF0D;6x+8y+24=0" role="presentation">C2:x2+y2-6x+8y+24=0 上,则 (  )
 A.|PQ|" role="presentation">|PQ| 的最小值为 0" role="presentation">0
 B.|PQ|" role="presentation">|PQ| 的最大值为 7" role="presentation">7
 C.两个圆心所在的直线斜率为 &#x2212;43" role="presentation">−43
 D.两个圆相交弦所在直线的方程为 6x&#xFF0D;8y&#xFF0D;25=0" role="presentation">6x-8y-25=0

4. 圆 C1:x2+y2+4x&#x2212;4y&#x2212;5=0" role="presentation">C1:x2+y2+4x−4y−5=0,圆 C2:x2+y2&#x2212;8x+4y+7=0" role="presentation">C2:x2+y2−8x+4y+7=0 的公切线条数是 &#x005F;" role="presentation">_.
 

5. 以 (3&#xFF0C;&#x2212;4)" role="presentation">(3,−4) 为圆心,且与圆 x2+y2=64" role="presentation">x2+y2=64 内切的圆的方程是 &#x005F;" role="presentation">_.
 

6. 已知点 A(&#x2212;1,0)" role="presentation">A(−1,0),B(1,2)" role="presentation">B(1,2),圆 C&#xFF1A;(x&#x2212;a)2+y2=25(a&gt;0)" role="presentation">C:(x−a)2+y2=25(a>0) 上存在唯一的点 P" role="presentation">P,使 PA2+PB2=12" role="presentation">PA2+PB2=12,则实数 a" role="presentation">a 的值是 &#x005F;" role="presentation">_.
 

7. 半径为 3" role="presentation">3 的圆与 x" role="presentation">x 轴相切,且与圆 x2+(y&#x2212;1)2=1" role="presentation">x2+(y−1)2=1 外切,求此圆的方程.
 

8. 求与圆 C:x2+y2&#x2212;2x=0" role="presentation">C:x2+y2−2x=0 外切且与直线 l:x+3y=0" role="presentation">l:x+3y=0 相切于点 M(3,&#x2212;3)" role="presentation">M(3,−3) 的圆的方程.
 

9. 已知圆 C1:x2+y2+2x&#x2212;6y+1=0" role="presentation">C1:x2+y2+2x−6y+1=0,圆 C2:x2+y2&#x2212;4x+2y&#x2212;11=0" role="presentation">C2:x2+y2−4x+2y−11=0.求两圆的公共弦所在直线的方程及公共弦长.
 

参考答案

答案 D" role="presentation">D
解析 圆 x2+y2=2" role="presentation">x2+y2=2 的圆心 (0&#xFF0C;0)" role="presentation">(0,0),半径为 R=2" role="presentation">R=2,
圆 x2+y2+4y+3=0" role="presentation">x2+y2+4y+3=0 化为标准方程得: x2+(y+2)2=1" role="presentation">x2+(y+2)2=1,
故圆心坐标 (0&#xFF0C;&#x2212;2)" role="presentation">(0,−2),半径为 r=1" role="presentation">r=1,
&#x2235;" role="presentation">∵ 圆心之间的距离 d=2" role="presentation">d=2,R+r=1+2&gt;2" role="presentation">R+r=1+2>2,R&#x2212;r=2&#x2212;1&lt;2" role="presentation">R−r=2−1<2
&#x2234;R&#x2212;r&lt;d&lt;R+r" role="presentation">∴R−r<d<R+r,
则两圆的位置关系是相交.
故选:D" role="presentation">D.

答案 D" role="presentation">D
解析 圆 (x&#x2212;1)2+(y+3)2=r2" role="presentation">(x−1)2+(y+3)2=r2 与 x2+y2=16" role="presentation">x2+y2=16,
所以两圆的圆心分别为 (1,&#x2212;3)" role="presentation">(1,−3),(0,0)" role="presentation">(0,0),半径分别为 r" role="presentation">r,4" role="presentation">4,
则两圆的圆心距为 (1&#x2212;0)2+(&#x2212;3&#x2212;0)2=10" role="presentation">(1−0)2+(−3−0)2=10,
因为 r&#xFF1E;0" role="presentation">r>0,
所以 r+4&gt;10" role="presentation">r+4>10,
则两圆不可能是外切和相离.
故选:D" role="presentation">D.

答案 BC" role="presentation">BC
解析 根据题意,圆 C1&#xFF1A;x2+y2=1" role="presentation">C1:x2+y2=1,其圆心 C1(0,0)" role="presentation">C1(0,0),半径 R=1" role="presentation">R=1,
圆 C2&#xFF1A;x2+y2&#xFF0D;6x+8y+24=0" role="presentation">C2:x2+y2-6x+8y+24=0,其圆心 C2(3,&#xFF0D;4)" role="presentation">C2(3,-4),半径 r=1" role="presentation">r=1,
圆心距 |C1C2|=16+9=5" role="presentation">|C1C2|=16+9=5,
则 |PO|" role="presentation">|PO| 的最小值为 |C1C2|&#x2212;R&#x2212;r=3" role="presentation">|C1C2|−R−r=3,最大值为 |C1C2|+R+r=7" role="presentation">|C1C2|+R+r=7,
故 A" role="presentation">A 错误,B" role="presentation">B 正确;
对于 C" role="presentation">C,圆心 C1(0,0)" role="presentation">C1(0,0),圆心 C2(3,&#x2212;4)" role="presentation">C2(3,−4),
则两个圆心所在的直线斜率 k=&#x2212;4&#x2212;03&#x2212;0=&#x2212;43" role="presentation">k=−4−03−0=−43,C" role="presentation">C 错误,
对于 D" role="presentation">D,两圆圆心距 |C1C2|=5" role="presentation">|C1C2|=5,有 |C1C2|&gt;R+r=2" role="presentation">|C1C2|>R+r=2,
两圆外切,不存在公共弦,D" role="presentation">D 错误.
故选:BC" role="presentation">BC.

答案 3" role="presentation">3
解析 圆 C1:(x+2)2+(y&#x2212;2)2=13" role="presentation">C1:(x+2)2+(y−2)2=13,圆 C2:(x&#x2212;4)2+(y+2)2=13" role="presentation">C2:(x−4)2+(y+2)2=13,
因此两圆的圆心坐标分别为 C1(&#x2212;2,2)" role="presentation">C1(−2,2),C2(4,&#x2212;2)" role="presentation">C2(4,−2),两圆的半径 r1=r2=13" role="presentation">r1=r2=13.
圆心距 |C1C2|=(&#x2212;2&#x2212;4)2+(2+2)2=213=r1+r2" role="presentation">|C1C2|=(−2−4)2+(2+2)2=213=r1+r2,
&#x2234;" role="presentation">∴ 两圆外切,有 3" role="presentation">3 条公切线.

答案 (x&#x2212;3)2+(y+4)2=9" role="presentation">(x−3)2+(y+4)2=9 或 (x&#x2212;3)2+(y+4)2=169" role="presentation">(x−3)2+(y+4)2=169
解析 设所求的圆的方程为 (x&#x2212;3)2+(y+4)2=r2&#xFF0C;r&gt;0" role="presentation">(x−3)2+(y+4)2=r2,r>0,
则由题意可得 9+16=|8&#x2212;r|" role="presentation">9+16=|8−r|,求得 r=3" role="presentation">r=3,或 r=13" role="presentation">r=13,
故所求的圆的方程为 (x&#x2212;3)2+(y+4)2=9" role="presentation">(x−3)2+(y+4)2=9 或 (x&#x2212;3)2+(y+4)2=169" role="presentation">(x−3)2+(y+4)2=169,
故答案为: (x&#x2212;3)2+(y+4)2=9" role="presentation">(x−3)2+(y+4)2=9 或 (x&#x2212;3)2+(y+4)2=169" role="presentation">(x−3)2+(y+4)2=169.

答案 &#x00B1;43" role="presentation">±43 或 &#x00B1;22" role="presentation">±22
解析 设 P(x,y)" role="presentation">P(x,y),则 PA2=(x+1)2+(y&#x2212;0)2=x2+y2+2x+1" role="presentation">PA2=(x+1)2+(y−0)2=x2+y2+2x+1,
PB2=(x&#x2212;1)2+(y&#x2212;2)2=x2+y2&#x2212;2x&#x2212;4y+5" role="presentation">PB2=(x−1)2+(y−2)2=x2+y2−2x−4y+5,
&#x2235;PA2+PB2=12" role="presentation">∵PA2+PB2=12,
&#x2234;x2+y2&#x2212;2y&#xFF0D;3=0" role="presentation">∴x2+y2−2y-3=0,即 x2+(y&#x2212;1)2=4" role="presentation">x2+(y−1)2=4;
&#x2234;P" role="presentation">∴P 点轨迹方程为 x2+(y&#x2212;1)2=4" role="presentation">x2+(y−1)2=4.
又圆 C" role="presentation">C 上存在唯一的点 P 符合题意,
&#x2234;" role="presentation">∴ 两圆相切,
&#x2234;a2+1=7" role="presentation">∴a2+1=7 或 a2+1=3" role="presentation">a2+1=3,解得 a=&#x00B1;43" role="presentation">a=±43 或 a=&#x00B1;22" role="presentation">a=±22,
&#x2234;" role="presentation">∴ 实数 a" role="presentation">a 的值为 &#x00B1;43" role="presentation">±43 或 &#x00B1;22" role="presentation">±22.

答案 (x&#x00B1;23)2+(y&#x2212;3)2=9" role="presentation">(x±23)2+(y−3)2=9 或 x2+(y+3)2=9" role="presentation">x2+(y+3)2=9
解析 因为所求圆的半径为 3" role="presentation">3 且与 x" role="presentation">x 轴相切,
所以设圆心坐标为 (a,&#xFF0D;3)" role="presentation">(a,-3) 或 (a,3)" role="presentation">(a,3).
又因为所求圆与圆 x2+(y&#x2212;1)2=1" role="presentation">x2+(y−1)2=1 外切,
所以 a2+4=4" role="presentation">a2+4=4 或 a2+16=4" role="presentation">a2+16=4,即 a=&#x00B1;23" role="presentation">a=±23 或 a=0" role="presentation">a=0.
所以所求圆的方程为 (x&#x00B1;23)2+(y&#x2212;3)2=9" role="presentation">(x±23)2+(y−3)2=9 或 x2+(y+3)2=9" role="presentation">x2+(y+3)2=9.

答案 (x&#x2212;4)2+y2=4" role="presentation">(x−4)2+y2=4
解析 圆 C" role="presentation">C 的方程可化为 (x&#x2212;1)2+y2=1" role="presentation">(x−1)2+y2=1,圆心 C(1,0)" role="presentation">C(1,0),半径为 1" role="presentation">1.
设所求圆的方程为 (x&#x2212;a)2+(y&#x2212;b)2=r2(r&gt;0)" role="presentation">(x−a)2+(y−b)2=r2(r>0),
由题意可得 {(a&#x2212;1)2+b2=r+1b+3a&#x2212;3&#x00D7;(&#x2212;33)=&#x2212;1|a+3b|2=r" role="presentation">{(a−1)2+b2=r+1b+3a−3×(−33)=−1|a+3b|2=r,解得 {a=4b=0r=2" role="presentation">{a=4b=0r=2,
所以所求圆的方程为 (x&#x2212;4)2+y2=4" role="presentation">(x−4)2+y2=4.

答案 245" role="presentation">245
解析 联立方程组 {x2+y2+2x&#x2212;6y+1=0(1)x2+y2&#x2212;4x+2y&#x2212;11=0(2)" role="presentation">{x2+y2+2x−6y+1=0(1)x2+y2−4x+2y−11=0(2)
①-②得 3x&#x2212;4y+6=0" role="presentation">3x−4y+6=0.
&#x2234;3x&#x2212;4y+6=0" role="presentation">∴3x−4y+6=0 即为两圆公共弦所在直线的方程.
易知圆 C1" role="presentation">C1 的圆 心 (&#x2212;1,3)" role="presentation">(−1,3),半径 r=3" role="presentation">r=3.
又 C1" role="presentation">C1 到直线 AB" role="presentation">AB 的距离为 d=|&#x2212;1&#x00D7;3&#x2212;4&#x00D7;3+6|32+42=95" role="presentation">d=|−1×3−4×3+6|32+42=95,
&#x2234;|AB|=2r2&#x2212;d2=232&#x2212;(95)2=245" role="presentation">∴|AB|=2r2−d2=232−(95)2=245,
即两圆的公共弦长为 245" role="presentation">245.
 

【B组---提高题】

1.(多选) 已知点 Q(4,0)" role="presentation">Q(4,0),过圆 (x&#x2212;4)2+y2=16" role="presentation">(x−4)2+y2=16 上的一动点 P" role="presentation">P 作圆 (x&#x2212;4)2+y2=4" role="presentation">(x−4)2+y2=4 的两条切线 PA" role="presentation">PA、PB" role="presentation">PB,切点分别为 A" role="presentation">A、B" role="presentation">B,两个切点 A" role="presentation">A、B" role="presentation">B 之间的线段 AB" role="presentation">AB 称为切点弦.则下列结论正确的是 (  )
 A.PQ&#x22A5;AB" role="presentation">PQ⊥AB " role="presentation"> B. |PA|=23" role="presentation">|PA|=23 " role="presentation"> C.|AB|=3" role="presentation">|AB|=3" role="presentation"> D.四边形 APBQ" role="presentation">APBQ 的面积为 43" role="presentation">43
 

2. 如图,设圆 C1&#xFF1A;(x&#x2212;5)2+(y+2)2=4" role="presentation">C1:(x−5)2+(y+2)2=4,圆 C2&#xFF1A;(x&#x2212;7)2+(y+1)2=25" role="presentation">C2:(x−7)2+(y+1)2=25,点 A" role="presentation">A、B" role="presentation">B 分别是圆 C1" role="presentation">C1,C2" role="presentation">C2 上的动点,P" role="presentation">P 为直线 y=x" role="presentation">y=x 上的动点,则 |PA|+|PB|" role="presentation">|PA|+|PB| 的最小值为 &#x005F;" role="presentation">_ .

 

参考答案

答案 ABD" role="presentation">ABD
解析 因为 Q(4,0)" role="presentation">Q(4,0) 为两已知圆的圆心,由几何性质可知 |PA|=|PB|" role="presentation">|PA|=|PB|,|QA|=|QB|" role="presentation">|QA|=|QB|,
所以 PQ&#x22A5;AB" role="presentation">PQ⊥AB,故 A" role="presentation">A 正确;
因为 |PQ|=4" role="presentation">|PQ|=4,|AQ|=|BQ|=2" role="presentation">|AQ|=|BQ|=2,
所以 |PB|=|PA|=|QP|2&#x2212;|QA|2=23" role="presentation">|PB|=|PA|=|QP|2−|QA|2=23,故 B" role="presentation">B 正确;
因为 sin&#x2061;&#x2220;APQ=|AQ||PQ|=12" role="presentation">sin⁡∠APQ=|AQ||PQ|=12,
又 &#x2220;APQ" role="presentation">∠APQ 为锐角,所以 &#x2220;APQ=30&#xB0;" role="presentation">∠APQ=30°,同理可得 &#x2220;BPQ=30&#xB0;" role="presentation">∠BPQ=30°,
所以 &#x2220;APB=60&#xB0;" role="presentation">∠APB=60°,则 &#x25B3;APB" role="presentation">△APB 为等边三角形,所以 |AB|=23" role="presentation">|AB|=23,
S&#x25B3;APBQ=2S&#x25B3;APQ=|PA|&#x22C5;|AQ|=43" role="presentation">S△APBQ=2S△APQ=|PA|⋅|AQ|=43,故 C" role="presentation">C 错误,D" role="presentation">D 正确,
故选:ABD" role="presentation">ABD. 答案 313&#x2212;7" role="presentation">313−7
解析 依题意可知圆 C1" role="presentation">C1 的圆心 (5,&#xFF0D;2)" role="presentation">(5,-2),r=2" role="presentation">r=2,圆 C2" role="presentation">C2 的圆心 (7,&#xFF0D;1)" role="presentation">(7,-1),R=5" role="presentation">R=5,如图所示:

对于直线 y=x" role="presentation">y=x 上的任一点 P" role="presentation">P,由图象可知,要使 |PA|+|PB|" role="presentation">|PA|+|PB| 的得最小值,
则问题可转化为求 |PC1|+|PC2|&#xFF0D;R&#xFF0D;r=|PC1|+|PC2|&#xFF0D;7" role="presentation">|PC1|+|PC2|-R-r=|PC1|+|PC2|-7 的最小值,
即可看作直线 y=x" role="presentation">y=x 上一点到两定点距离之和的最小值减去 7" role="presentation">7,
由平面几何的知识易知当 C1" role="presentation">C1 关于直线 y=x" role="presentation">y=x 对称的点为 C1&#x2032;(&#xFF0D;2,5)" role="presentation">C1′(-2,5),
与 P" role="presentation">P、C2" role="presentation">C2 共线时,|PC1|+|PC2|" role="presentation">|PC1|+|PC2| 的最小值,
取得最小值,即直线 y=x" role="presentation">y=x 上一点到两定点距离之和取得最小值为 |CC2|=313" role="presentation">|CC2|=313,
&#x2234;|PA|+|PB|" role="presentation">∴|PA|+|PB| 的最小值 [=|PC1|+|PC2|&#x2212;7=313&#x2212;7" role="presentation">[=|PC1|+|PC2|−7=313−7.
 

【C组---拓展题】

1. 两圆 x2+y2+2ax+2ay+2a2&#x2212;1=0" role="presentation">x2+y2+2ax+2ay+2a2−1=0 和 x2+y2+2bx+2by+2b2&#x2212;2=0" role="presentation">x2+y2+2bx+2by+2b2−2=0 的公共弦中,最长的弦等于 &#x005F;" role="presentation">_.
 

参考答案

答案 2" role="presentation">2
解析 将两圆方程相减得公共弦所在直线方程: (2a&#x2212;2b)x+(2a&#x2212;2b)y+2a2&#x2212;2b2+1=0" role="presentation">(2a−2b)x+(2a−2b)y+2a2−2b2+1=0,
x2+y2+2ax+2ay+2a2&#x2212;1=0" role="presentation">x2+y2+2ax+2ay+2a2−1=0 化为标准方程为 (x+a)2+(y+a)2=1" role="presentation">(x+a)2+(y+a)2=1,
其圆心为 (&#x2212;a,&#x2212;a)" role="presentation">(−a,−a),半径为 1" role="presentation">1,
圆心到公共弦所在直线的距离为 d=|(2a&#x2212;2b)(&#x2212;a)&#x00D7;2+2a2&#x2212;2b2+1|22|a&#x2212;b|=|1&#x2212;2(a&#x2212;b)2|22|a&#x2212;b|" role="presentation">d=|(2a−2b)(−a)×2+2a2−2b2+1|22|a−b|=|1−2(a−b)2|22|a−b|
所以公共弦长为 21&#x2212;d2=21&#x2212;(1&#x2212;2(a&#x2212;b)222|a&#x2212;b|)2" role="presentation">21−d2=21−(1−2(a−b)222|a−b|)2=21&#x2212;((a&#x2212;b)22+18(a&#x2212;b)2&#x2212;12)" role="presentation">=21−((a−b)22+18(a−b)2−12)
&#x2264;21&#x2212;(2116&#x2212;12)=2" role="presentation">≤21−(2116−12)=2,
故答案为:2" role="presentation">2.

网址:2.5.2 圆与圆的位置关系 https://mxgxt.com/news/view/457244

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