题目内容
如图,平面直角坐标系中,⊙P与x轴分别交于A、B两点,点P的坐标为(3,-1),AB=2
.若将⊙P向上平移,则⊙P与x轴相切时点P坐标为( )
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3 |
A.(3,2) | B.(3,3) | C.(3,4) | D.(3,5) |
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当P移到P′点时,⊙P与x轴相切,
过P作直径MN⊥AB与D,连接AP,
由垂径定理得:AD=BD=
1 |
2 |
3 |
∵DP=|-1|=1,
由勾股定理得:AP=
AD2+PD2 |
∴PP′=2+1=3,
∵P(3,-1),
∴P′的坐标是(3,2),
故选A.
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