請教A.Math！我補習阿ｓｉｒ都５識ｅ條！

The the request slope of equation be M.
Since this line perpendicular to 12x-5y-53=0,
so, M*12/5=-1
i.e., M= -5/12

For the circle, x^2+y^2+16x-8y-89=0
differentiate both sides with respect to x:
2x + 2y dy/dx +16 -8 dy/dx =0

dy/dx = -(x+8) / (y-4)
when the request line touch the circle,
dy/dx = M
i.e., -(x+8) / (y-4) = -5/12
ans: 12x - 5y + 116 = 0

2008-05-18 02:10:07 補充：
都得，

let the request equation be (y-y1)/(x-x1)=M = -5/12

The circle: centre(-8, 4), radius = 13

one of the radius is perpendicular to the request line,
this radius is (y-4)/(x+8) =12/5 and it touch the circle at the point which same as the request line touchs.

2008-05-18 02:13:46 補充：
so,
radius: 12x - 5y + 116 =0
circle: x^2+y^2+16x-8y-89=0
solve for x and y, will have 2 answers, so we have 2 request lines.

2008-05-18 02:29:40 補充：
x^2 + (144x^2 + 2784x + 13456)/25 + 16x -(96x + 928)/5 -89 =0
169x^2 + 2704x +6591=0
x= -3 or -13
for x =-3, y= 16
for x = -13, y= -8

the request line:
y-16/x+3 =-5/12
or
y+ 8/x+13 =-5/12

2008-05-18 02:31:58 補充：
ANSWERS:
5x+12y-177=0
or
5x+12y+161=0

2008-05-20 12:40:52 補充：
係喎，都幾好，不過驚佢未學。

The current answer is correct. However, we can think of it geometrically.
Circle: r=13, centre (-8,4).
Line=5x+12y+k=0
Distance from centre:
(5(-8)+12(+4)+k)/sqrt(5^2+12^2)=13
or -40+48+k=+/- 169, hence
k=-177 or 161