If the endpoints of the diameter of a circle are (−14, 4) and (−2, 0), what is the standard form equation of the circle?

First calculate the  radius  = 1/2 the diameter =  [sqrt  sqr160(4-0)^2 + (-14-(-2))^2] / 2

=  (sqrt 160 ) / 2  so r^2 =  160 /4  = 40

center = midpoint of the diameter  = (-14-2)/ 2  ,    (4+0)/2  =  (-8, 2)

so required equation is  (x + 8)^2 + (y - 2)^2 = 40

Kis the midpoint of bc