Equations of circle

The Q is solved as follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths285.jpg?t=1214200543


2008-06-23 16:13:00 補充：
It is alright to use calculus to find the slopes, but they can actually be found in a less complicated way and it is advisable to solve questions with simple methods.

2008-06-23 16:44:05 補充：
I have added in more explanation, plx press reload for it.

a)
x^2 +y^2 - 4x +8y +10 = 0 ----------(1)
x -2y - 5 = 0 ---------------------------(2)
from (2)
x = 2y +5 ------------------------------(3)
put (3) into (1)
(2y +5)^2 + y^2 - 4(2y+5) +8y +10 = 0
y^2 + 4y +3 = 0
y = -1 or y = -3
let f(y) = 2y +5
f(-1) = 3
f(-3) = -1
the points of intersection of C and L are (3,-1) and (-1,-3)

b)
x^2 +y^2 -4x +8y +10 = 0
2x + 2y*dy/dx -4 +8*dy/dx = 0
(dy/dx) = (4 - 2x)/(2y+8)
slope of the circle at point A(3,-1) are -1/3
slope of OA are (-1-0)/(3-0) = -1/3
slope of the circle at point B(-1/-3) are 3
slope of OB are (-3-0)/(-1-0) = 3

ci)
equation of OA: y/x = -1/3
y = -1/3*x
equation of OB: y/x = 3
y = 3x

cii)
d(OA)/dx = -1/3
d(OB)/dx = 3
d(OA)/dx*d(OB)/dx = -1
so they are at right angles