Question 1044110
The centre of a circle has coordinates (0,0) One end of a diameter is located at (7,-2)

a) What are the coordinates of the other endpoint of this diameter? 

b) What is the equation of the circle? 

many ways to tackle this.
{{{c(0,0)}}},{{{D(7,-2)}}}
by inspection the other end of the diameter is located at {{{D(-7,2)}}}.
use midpoint formula to verify
({{{0,0)}}}={{{(a+7)/(2)+(b+-2)/2}}}
means
{{{(0)}}}={{{(a+7)/(2)}}},{{{a+7=0}}},{{{a=-7}}}
{{{0}}}={{{(b+-2)/2}}},{{{b-2=0}}},{{{b=2}}} therefore {{{D(-7,2)}}}

two ways 
find the distance from end of the diamater to other end to get diameter ,divide by 2 to  get radius.using (7, -2) and (-7, 2)

{{{D }}} = {{{( sqrt ( 212))}}}
{{{D }}} = {{{( sqrt ( 212))}}}
{{{D }}} = 2{{{( sqrt ( 53))}}}
half of it is the radius or {{{R }}} = {{{( sqrt ( 53))}}}

same can be found using origin and one end of the diameter to get radius.
distance from one end to the origin c(0,0) to get radius using ((0 ,0))and ((7 ,-2))
{{{R}}} = {{{( sqrt ( 53))}}}

the equation is
{{{x^2+y^2}}}={{{r^2}}}
{{{x^2+y^2}}}={{{(sqrt(53))^2

{{{x^2+y^2}}}={{{53}}}