题目内容
给定圆C:x2+y2=4,过点P(1,0)作两条互相垂直的直线与C分别交于A、B和M,N,则![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_ST/1.png)
【答案】分析:由圆C的方程找出圆心C的坐标和半径r,设出直线AB的斜率为k,根据两直线垂直时斜率的乘积为-1,得到直线MN的斜率为-
,由两直线都过P点,进而分别表示出两直线的方程,利用点到直线的距离公式分别求出圆心到两直线的距离d1和d2,由垂径定理得到垂足为中点,由弦心距,半径,利用勾股定理求出弦的一半,进而表示出|AB|和|MN|,得出|AB|2+|MN|2的值为定值,再表示出|MN|•|AB|,变形后求出|MN|•|AB|的最小值,把所求的式子通分后,将求出的|AB|2+|MN|2的值及|MN|•|AB|的最小值代入,即可求出所求式子的最大值.
解答:解:由圆的方程x2+y2=4,得到圆心坐标为(0,0),半径r=2,
设直线AB的方程为:y=k(x-1),即kx-y-k=0,
则直线MN的方程为:y=-
(x-1),即x+ky-1=0,
∴圆心到直线AB的距离d1=
,到直线MN的距离d2=
,
∴|AB|=2
=2
,|MN|=2
=2
,
∵|MN|•|AB|=4![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/8.png)
=4
=4
≥4
=8
,
∴(|MN|•|AB|)min=8
,
∵|AB|2+|MN|2=4(
+
)=
=28,
∴
+
=
=
,
当(|MN|•|AB|)min=8
时,
则(
+
)max=
=
.
故答案为:![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/26.png)
点评:此题考查了直线与圆相交的性质,涉及的知识有:垂径定理,勾股定理,两直线垂直时斜率满足的关系,直线的一般式方程,以及圆的标准方程,其中得出|AB|2+|MN|2的值为定值,同时求出|MN|•|AB|的最小值是解本题的关键.
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/0.png)
解答:解:由圆的方程x2+y2=4,得到圆心坐标为(0,0),半径r=2,
设直线AB的方程为:y=k(x-1),即kx-y-k=0,
则直线MN的方程为:y=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/1.png)
∴圆心到直线AB的距离d1=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/3.png)
∴|AB|=2
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/7.png)
∵|MN|•|AB|=4
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/8.png)
=4
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/12.png)
∴(|MN|•|AB|)min=8
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/13.png)
∵|AB|2+|MN|2=4(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/16.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/17.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/19.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/20.png)
当(|MN|•|AB|)min=8
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/21.png)
则(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/23.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/24.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/25.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184920607114890/SYS201310241849206071148013_DA/26.png)
点评:此题考查了直线与圆相交的性质,涉及的知识有:垂径定理,勾股定理,两直线垂直时斜率满足的关系,直线的一般式方程,以及圆的标准方程,其中得出|AB|2+|MN|2的值为定值,同时求出|MN|•|AB|的最小值是解本题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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