求过点A(2,3)且与圆(x-1)²+(y-1)²=1相切的切线方程
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求过点A(2,3)且与圆(x-1)²+(y-1)²=1相切的切线方程
求过点A(2,3)且与圆(x-1)²+(y-1)²=1相切的切线方程
求过点A(2,3)且与圆(x-1)²+(y-1)²=1相切的切线方程
切线有两条:
一、平行于y轴的直线:
x=2
二、设直线斜率为k,
设直线的方程为:y=kx+b
带入点A(2,3) 3=2k+b
即 b=3-2k
所以该直线的方程为:y=kx+3-2k
因为直线与圆相切,所以圆心到直线的距离为半径的长,即为1.
|k+3-2k-1|/根号(1+k^2)=1
(2-k)^2=1+k^2 k=3/4
所以:y=3/4x+3/2
过点P(2,3)且与圆(x-1)^2+(y-1)^2=1相切的直线中有一条平行于y轴
直线方程为x=3
设另一条的斜率为k,显然k>0
y-3=k(x-3)
y-kx+2k-3=0
圆心(0,0)到直线的距离为半径2
|2k-3|/√(1+k^2)=2
k=5/12
直线方程为
y-5x/12-13/6=0
12y-5x-26=0