A maths (circles)

The equation of the tangents to the circle at P (x0,y0) is
x0x+y0y+d(x0+x)/2+e(y0+y)/2+f=0
The equation of the tangents to the circle at A(6,0) is
6x-2(x+6)+(1/2)y-12=0
12x-4x-24+y-24=0
8x+y-48=0
y=48-8x...(1)
The equation of the tangents to the circle at B(0,-4) is
-4y-2x+1/2(y-4)-12=0
-8y-4x+y-4-24=0
-7y-4x-28=0
7x+4y+28=0...(2)
(1) sub into (2)
7x+4(48-8x)+28=0
7x+192-32x+28=0
25x=220
x=8.8
y=48-8x=-22.4
the coordinates of the point of intersection of the tangents to the circle at A and B is (8.8, -22.4)

2007-01-13 22:55:21 補充：
做錯應是-7y-4x-28=04x+7y+28=0...(2)(1) sub into (2)4x+7(48-8x)+28=04x+336-56x+28=0-52x=-364x=7y=48-8x=-8the coordinates of the point of intersection of the tangents to the circle at A and B is (7, -8)上面個位切線方程做錯

Method I:
By using the formula of the equation of the tangent to the circle
ax+by+(a+x)/2+(b+y)/2+F=0 (where a is the x-coordinate and b is the y-coordinate of the pt. of intersection and F is the constant term of the equation of the circle)

the equation of the tangent at A is
6x+0y+(6+x)/2+(0+y)/2-12=0
13x+y-9=0

the equation of the tangent at B is
0x-4y+(0+x)/2+(y-4)/2-12=0
x-7y-28=0
Method II: let the tangent at A be y=m(x-6), where m is the slope
sub y=m(x-6) into the equation of the circle
x^2+m^2(x-6)^2-4x+m(x-6)-12=0
by usint dicriminant=0, u can find the value of m and finally get the answer.
the tangent at B is similar to that one. But i think the method I is better