idglobe
幼苗
共回答了15个问题采纳率:86.7% 举报
(1)x-2y+2±
![](https://img.yulucn.com/upload/2/de/2debc318d26c01af40fbb0ba7331f747_thumb.jpg)
=0
(2)
(1)圆C的方程为x
2 +(y-1)
2 =1,其圆心为C(0,1),半径r=1.
由题意可设直线l′的方程为x-2y+m=0.
由直线与圆相切可得C到直线l′的距离d=r,即
![](https://img.yulucn.com/upload/1/b2/1b22a8e31666e4dc05881b43047183a1_thumb.jpg)
=1,解得m=2±
![](https://img.yulucn.com/upload/2/de/2debc318d26c01af40fbb0ba7331f747_thumb.jpg)
.
故直线l′的方程为x-2y+2±
![](https://img.yulucn.com/upload/2/de/2debc318d26c01af40fbb0ba7331f747_thumb.jpg)
=0.
(2)结合图形可知:|PT|=
![](https://img.yulucn.com/upload/a/46/a4634d26d34f4543072b3cfd55abddee_thumb.jpg)
=
![](https://img.yulucn.com/upload/6/2b/62b0bf9484cfb2ebea947c50bb4b2b60_thumb.jpg)
.故当|PC|最小时,|PT|有最小值.
易知当PC⊥l时,|PC|取得最小值,且最小值即为C到直线l的距离,得|PC|
min =
![](https://img.yulucn.com/upload/2/14/2148889a4654f942dbac1de22cd29b58_thumb.jpg)
.
所以|PT|
min =
![](https://img.yulucn.com/upload/9/f7/9f74994782d9e36c37f730b92de0dac4_thumb.jpg)
=
![](https://img.yulucn.com/upload/f/27/f2734bb941d59ff1ba9cc4901ae12195_thumb.jpg)
.
1年前
7