Question 705215
If (0,0) and (4,-6) are endpoints of a diameter,
the midpoint of that diameter is the center of the circle.
Its coordinates are found by averaging the coordinates of those endpoints.
The x-coordinate is {{{(0+4)/2=highlight(2)}}}.
The y-coordinate is {{{(0+(-6))/2=highlight(-3)}}}
So, we have the center, C(2,-3).
 
A picture is work a thousand words:
{{{drawing(300,300,-2,6,-7,1,
grid(1),
red(circle(2,-3,sqrt(13))),
blue(circle(0,0,0.1)),blue(circle(4,-6,0.1)),
blue(line(0,0,4,-6)),blue(circle(2,-3,0.1)),
locate(2.1,-3,blue(C(2,-3))),
green(triangle(0.02,0.02,1.98,-3,0.02,-3)),
green(rectangle(0.02,-3,0.22,-2.8))
)}}} The radius, {{{R}}}, is the length of the segment form (0,0) to C(2,-3).
That segment is the hypotenuse of a right triangle (drawn in green).
We can calculate {{{R^2}}} based on the Pythagorean theorem
{{{R^2=2^2+3^2}}} --> {{{R^2=4+9}}} --> {{{R^2=highlight(13)}}}
Any point (x,y) in the circle is at the same distance {{{R}}} from the center, C.
As a consequence
{{{(x-2)^2+(y-(-3))^2=R^2}}} --> {{{highlight((x-2)^2+(y+3)^2=13)}}}