equation of circle(2)

(x^2)+(y^2)-6x+8y-75=0
The centre of the circle = ( 3,-4 )
The radius of the circle = (1/2) [ (-6)^2 + (8)^2 - 4(-75) ] ^ 1/2
r = 10

Let the mid pt of PQ be (x,y).
x = ( -3+9 )/2 = 3
y = ( 4 + (-12) )/2 = -4
So the mid pt of the circle is ( 3,-4 )
PQ^2 = [-3-9]^2 + [4-(-12)]^2
PQ^2 = 400
PQ = 20 = 2r

Since PQ passes through the centre and the PQ = 2r ,
so the line segment joining P(-3,4) and Q(9,-12) is a diameter of the circle .

2011-06-01 19:47:46 補充：
If there are any redundant steps in my answer , welcome to point out for me ^^