(1)∵点A(4,-3)在圆x2+y2=25上,
圆心:O(0,0),半径r=5,
∴kOA=-[3/4],∴切线方程过A(4,-3),斜率k=-[1
kOA=
4/3],
∴过点A(4,-3)的切线方程为y+3=[4/3(x?4),
整理,得4x-3y-25=0.
(2)设过点B的切线方程为y-2=k(x+5),即kx-y+5k+2=0,
则
|5k+2|
k2+1=5,解得k=
21
20].
∴过点B的切线方程为
21
20x?y+
105
20+2=0,
整理,得21x-20y+145=0.
当过点B的切线的斜率不存在时,切线方程为x=-5,成立.
综上,过点B的切线方程为21x-20y+145=0或x=-5.