题目内容
已知圆P:(x-m)2+(y-n)2=4与y轴交于A、B两点,且|
+
|=
,则|AB|=______.
| PA |
| PB |
| 10 |
设点E是AB的中点,连结PE,则
∵PE是△PAB的中线,
∴向量
| PE |
| 1 |
| 2 |
| PA |
| PB |
又∵|
| PA |
| PB |
| 10 |
| PE |
| ||
| 2 |
∵⊙P中,E是弦AB的中点
∴PE⊥AB,可得|AE|=
|
4-(
|
| ||
| 2 |
因此,|AB|=2|AE|=
| 6 |
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题目内容
| PA |
| PB |
| 10 |
| PE |
| 1 |
| 2 |
| PA |
| PB |
| PA |
| PB |
| 10 |
| PE |
| ||
| 2 |
|
4-(
|
| ||
| 2 |
| 6 |