Question 1182918
The diameter of Circle q terminates on the circumference of the circle at 

({{{3}}},{{{0}}}) and ({{{-4}}},{{{0}}})  

use given points and distance formula to find diameter

{{{d=sqrt((-4-3)^2+(0-0)^2)}}}
{{{d=sqrt((-7)^2)}}}
{{{d=sqrt(49)}}}
{{{d=7}}}

=> radius is {{{r=7/2}}}

now find coordinates of the center which is midpoint of diameter

({{{h}}},{{{k}}})=({{{(3-4)/2}}}, {{{0}}})
({{{h}}},{{{k}}})=({{{-1/2}}}, {{{0}}})

 the equation of the circle in standard form:

{{{(x-h)^2+(y-k)^2=r^2}}}
{{{(x-(-1/2))^2+(y-0)^2=(7/2)^2}}}
{{{(x+1/2)^2+y^2=49/4}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(3,0,.12), locate(3,0.5,p(3,0)),
circle(-4,0,.12), locate(-4,0.5,p(-4,0)),
circle(-1/2,0,.12), locate(-1/2,0.9,C(-1/2,0)),
graph( 600, 600, -10, 10, -10, 10,-sqrt(49/4-(x+1/2)^2) ,sqrt(49/4-(x+1/2)^2))) }}}