What is the equation of the circle that has a diameter with the end points located at (-5,-3) and (-11,-3)
a. (x-8)^2 + (y-3)^2=9
b. (x+5)^2+(y+3)^2=9
c. (x+8)^2+(y+3)^2=9
d. (x+8)^2+(y+3)^2=36

First find the radius which is half of the distance between the endpoints

r =  0.5 * ( 11-5) =  3

and the x coordinate of the center of the circle is  ( -11--5) / 2 =  -8
so center is at (-8,-3)

as equation of circle is (x -a)^2 + (y - b)^2 = r^2   where (a,b) = center

our circle will have the equation

(x + 8)^2 + (x + 3)^2 = 9

Step-by-step explanation: