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问题 解答题
已知椭圆C1：
x2
4
+y2=1和动圆C2：x2+y2=r2(r＞0)，直线l：y=kx+m与C1和C2分别有唯一的公共点A和B．
（I）求r的取值范围；
（II ）求|AB|的最大值，并求此时圆C2的方程．
答案

（Ⅰ）由

x2
4
+y2=1
y=kx+m
，得（1+4k2）x2+8kmx+4（m2-1）=0．

由于l与C1有唯一的公共点A，故△1=64k2m2-16（1+4k2）（m2-1）=0，

从而m2=1+4k2 ①

由

x2+y2=r2
y=kx+m
，得（1+k2）x2+2kmx+m2-r2=0．

由于l与C2有唯一的公共点B，故△2=4k2m2-4（1+k2）（m2-r2）=0，

从而m2=r2（1+k2） ②

由①、②得k2=

r2-1
4-r2
．

由k2≥0，得1≤r2＜4，所以r的取值范围是[1，2）．

（Ⅱ）设A（x1，y1），B（x2，y2），由（Ⅰ）的解答可知

x1=-

4km
1+4k2
=-
4k
m
，x2=-
km
1+k2
=-
kr2
m
．

|AB|2=（1+k2）（x2-x1）2=（1+k2）•

k2(4-r2)2
m2
=
1+k2
m2
•k2•（4-r2）2

=

1
r2
•
r2-1
4-r2
•（4-r2）2=
(r2-1)(4-r2)
r2
，

所以|AB|2=5-（r2+

4
r2
）（1≤r＜2）．

因为r2+

4
r2
≥2×2=4，当且仅当r=
2
时取等号，

所以当r=

2
时，|AB|取最大值1，此时C2的方程为x2+y2=2．

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填空题

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