Question

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

Answer 1

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