解
对于圆(x-a)^2+(y-b)^2=r^2,是圆x^2+y^2=r^2按向量(a,b)的平移,故可先求圆x^2+y^2=r^2与过圆外一定点的切线.
设圆O方程为x^2+y^2=r^2,定点P(x0,y0),切点Q(x1,y1).切线斜率k1,与切线垂直的半径斜率k2.
∴k1=-1/k2=-1/(y1/x1)=-x1/y1=(y1-y0)/(x1-x0),x1^2+y1^2=r^2.
整理得：x0x1+y0y1=r^2
当y0≠0,时y1=(r^2-x0x1)/y0.
∴x1^2+[(r^2-x0x1)/y0]^2=r^2
整理得：(x0^2+y0^2)x1^2-2x0r^2x1+r^4-y0^2r^2=0
解得：x1'={x0r^2+√[y0^2r^2(x0^2-r^2+y0^2)]}/(x0^2+y0^2)
x1''={x0r^2-√[y0^2r^2(x0^2-r^2+y0^2)]}/(x0^2+y0^2)
∴k1'=-x1'/y1
=y0x1'/(x0x1'-r^2)
=y0{x0r^2+√[y0^2r^2(x0^2-r^2+y0^2)]}/(x0^2+y0^2)/{x0{x0r^2+√[y0^2r^2(x0^2
r^2+y0^2)]}/(x0^2+y0^2)-r^2}
={x0y0r^2+y0√[y0^2r^2(x0^2-r^2+y0^2)]}/{x0√[y0^2r^2(x0^2-r^2+y0^2)]-y0^2r^2}
同理：k1''={x0y0r^2-y0√[y0^2r^2(x0^2-r^2+y0^2)]}/{-x0√[y0^2r^2(x0^2-r^2+y0^2)]-y0^2r^2}
当y0=0时也满足上式.
∴切线方程为：
y={{x0y0r^2+y0√[y0^2r^2(x0^2-r^2+y0^2)]}/{x0√[y0^2r^2(x0^2-r^2+y0^2)]-y0^2r^2}}(x-x0)+y0
或y={{x0y0r^2-y0√[y0^2r^2(x0^2-r^2+y0^2)]}/{-x0√[y0^2r^2(x0^2-r^2+y0^2)]-y0^2r^2}}(x-x0)+y0
再将(x0,y0)按向量(a,b)平移得:(x0+a,y0+b)代入上述切线方程即得圆(x-a)^2+(y-b)^2=r^2与过圆外一定点的切线方程
(一般题目给出圆外一定点(x0,y0)将它按(-a,-b)平移代入上式即得斜率,然后再求切线方程,)