Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$ Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$.

My Work:
$$x^2+y^2-6x+2y-8=0$$
Comparing the above equation with $x^2+y^2+2gx+2fy+c=0$, we get:
$$g=-3$$,
$$f=1$$
Now, 
$$\textrm {centre}=(-g,-f)$$
$$=(3,-1)$$
Search for "Completing the Square".

Solving your system $x^{2}+y^{2}−6x+2y−8=0$, you can add and substract some numbers to get the classical circle equation $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Well, you can check that your problem is equivalent to $$(x-3)^2+(y+1)^2=18$$ So your radius is $\sqrt{18}=3\sqrt{2}$, and your center $(3,-1)$

I dont know what you mean about "the equation of the diamater" but if you have a point $(a,b)$ the equation that passes through the point and the diamenter is:
$$y=\frac{b-3}{a+1}(x-a)+b$$
Complete the square

$$x^2 + y^2 -6x + 2y -8 = 0$$
$$x^2-6x + 9 - 9 +y^2+2y + 1-1-8= 0$$
$$(x-3)^2+(y+1)^2 = 18$$

Therefore, $18 = r^2,$ and diameter $= 6\sqrt 2$.