过圆O外一点M(a,b)向圆O:x方 y方=r方引两条切线,切点分别为A.B,求直线AB的方程

问题描述:

过圆O外一点M(a,b)向圆O:x方 y方=r方引两条切线,切点分别为A.B,求直线AB的方程
有四五种解法,能都写出来么
1个回答 分类:数学 2014-12-03

问题解答:

我来补答
1,
(x-a)^2+(y-b)^2=(a^2+b^2)-r^2.(1)
x^2+y^2=r^2.(2)
(1)-(2):
AB:
2,
y-b=k*(x-a).(1)
x^2+y^2=r^2.(2)
(1),(2):
(1+k^2)x^2+2k(b-ka)x+(b-ka)^2-r^2=0
[2k(b-ka)]^2-4*(1+k^2)[(b-ka)^2-r^2]=0
kA= kB=
xA=k(ka-b)/(1+k^2),yA=
xB= yB=
k(AB)=(yA-yB)/(xA-xB)
AB:
3,
y-b=k(x-a)
kx-y+b-ka=0
d=r,d^2=r^2
|b-ka|^2/(1+k^2)=r^2
kA= kB=
AB:
4,
[(y-b)/(x-a)]*(y/x)=-1.(1)
x^2+y^2=r^2.(2)
xA= yA=
xB= yB=
k(AB)=
AB:
5,
切点(x,y):
OA:y=kx.(1)
x^2+y^2=r^2.(2)
(1),(2):
xA= yA=
xB= yB=
(yA/xA)*[(yA-b)/(xA-a)]=-1
kA= xA= yA=
kB= xB= yB=
6,
P(a,b)
k(AB)=-1/k(OP)=-a/b
AB:y=(-a/b)x+m.(1)
x^2+y^2=r^2.(2)
xA= yA=
xB= yB=
[(yA-b)/xA-a)]*(yA/xA)=-1
m=
[(yB-b)/(xB-b)]*(yB/xB)=-1
m=
AB:
7,
PA^2=PB^2=OP^2-r^2=a^2+b^2-r^2=
|PA|=|PB|=
∠AOP=arcsin(PA/OP)
k(OA)=
OA:y=kx.(1)
x^2+y^2=r^2.(2)
(1),(2)
xA= yA=
xB= yB=
AB:
 
 
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