Question 1044110
({{{0}}}, {{{0}}}) is the midpoint of the diameter. Use the midpoint formula to determine the other end point.

{{{0=(x+7)/2}}}=>{{{x+7=0}}}=>{{{x=-7}}}
{{{0=(y+(-2))/2}}}=>{{{y-2=0}}}=>{{{y=2}}}
the other endpoint is at ({{{-7}}},{{{2}}})


b)
recall standard formula for a circle:
{{{(x-h)^2 + (y-k)^2 = r^2}}},  where {{{r}}} being the radius , {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the center

since  the center is at origin, {{{h=0}}} and {{{k=0}}} and your formula is

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


 the length of the radius is {{{1/2}}} of the distance between two endpoints 


{{{r=sqrt( (x1-x2)^2+(y1-y2)^2)/2}}}

{{{r=sqrt( (7-(-7))^2+(-2-2)^2)/2}}}

{{{r=sqrt( (7+7)^2+(-2-2)^2)/2}}}

{{{r=sqrt( 14^2+(-4)^2)/2}}}

{{{r=sqrt( 196+16)/2}}}

{{{r=sqrt( 212)/2}}}

{{{r=sqrt( 53*4)/2}}}

{{{r=2sqrt( 53)/2}}}

{{{r=sqrt( 53)}}}.......exact solution
so, your equation is:

{{{x^2 + y^2 = (sqrt( 53))^2}}}

{{{highlight(x^2 + y^2 = 53)}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(7,-2,.13),circle(-7,2,.13),
locate(7,-2,p(7,-2)),locate(-7,2,p(-7,2)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(-x^2+53),-sqrt(-x^2+53)) ) }}}