Question 1030760
{{{r^2=((-4-2)^2+(2-(-5))^2)}}}
{{{r^2=(36+49)}}}
{{{r^2=(85)}}}


{{{(x-2)^2+(y+5)^2=85}}}------but not really what you need.


The distance from center to any point on the circle is {{{sqrt(85)}}}.
The center, and the endpoints for any of the diameters of the circle, are collinear.


The line containing this diameter contains the points  (2,-5) and (-4,2).  Using point-slope form, the equation for this line is  {{{y+5=((2+5)/(-4-2))(x-2)}}}
{{{y+5=-(7/8)(x-2)}}}
{{{y=-(7/8)x+14/8-5}}}
{{{y=-(7/8)x+7/4-5}}}......
upon further thinking, this approach seems more complicated than necessary.




The center (2,-5)  is the MIDPOINT between the two endpoints for the diameter; and one endpoint is given to be  (-4,2).  The unknown endpoint is some ordered pair  (x,y).  Setup the midpoint formula for the circle's center and the two endpoints.


{{{system((x+(-4))/2=2,(y+2)/2=-5)}}}


{{{system(x-4=2*2,y+2=-10)}}}


{{{system(x=8,and,y=-12)}}}


The other diameter's endpoint is  (8,-12).