Find an equation of the circle whose endpoints of a diameter are (−3,6) and (5,−14).

Question

I tried to figure it out, but I'm not sure if my answer is correct.
I got..
(x-2)^2 + (y + 4)^2 = 116

Any help would be great. Thank you!

Answer 1

the centre must be the midpoint of the diameter which would give you (1, -4)
so the equation must be
(x-1)^2 + (y+4)^2 = r^2

sub in one of the points on the circle , I will use (-3,6)
(-4)^2 + (10)^2 = r^2
r^2 = 16+100= 116

so (x-1)^2 + (y+4)^2 = 116

check by subbing in the other endpoint, it works

You made one small error

Answer 2

Thank you very much your help. I appreciate it! :)