Question 941342
{{{x^2-4x+y^2+6y=14}}}
{{{x^2-4x+4+y^2+6y+9=14+4+9}}}
{{{(x-2)^2+(y+3)^2=27}}}


The LINE containing the diameter would contain the circle's center of (2,-3), AND also the origin (0,0).  You could put limitations on the domain for the line to achieve just the diameter segment; but in any case, you can find the line containing those two points.  The equation for the line is  {{{y=-(3/2)x}}}.


Find the endpoints of the diameter using the equation of the circle and the equation y=-(3/2)x.  Substitute for y in the circle equation, solve for x, and find the corresponding y value....  you have the endpoints of the diameter.