Math: Distance and Midpoint
posted by Anonymous .
Three points on the edge of a circle are (220, 220), (0, 0), and (200, 40), where each unit represents 1 foot. What is the diameter of the circle to the nearest 10 feet?
I know the answer is supposed to be 550 ft, but I keep coming up with the wrong answer. I found the midpoint of (220, 220) and (0, 0) which is (110, 110). Its perpendicular bisector is y = (1/2)x +165, if I did it correctly. I also found the midpoint of (0, 0) and (200, 40), which is (100, 20). I got y = 5x + 520 for its perpendicular bisect. Using those two equations, I found the intersect/center of the circle, (710/11, 2170/11). But when I use the distance formula with that point and (0, 0) to find the distance/radius, I get about 207.5, and the diameter about 415. What did I do wrong?

Alas, things have gone awry.
The slope of the line from (220,220) to (0,0) is 1. So, the slope of the perpendicular bisector (pb) is 1. The equation of the pb is thus
(y110) = (x+110)
y = x + 220
The slope of the line from (0,0) to (200,40) is 40/200 = 1/5. The pb slope is thus 5. The pb equation is
(y20) = 5(x100)
y = 5x +520
The two pb's intersect at (50,270)
so, it looks like the circle is
(x30)^2 + (y+30)^2 = 75400
sqrt(75400) = 274.59 or about 275
diameter would thus be 550. 
Oops. The equation is
(x50)^2 + (y+270)^2 = 75400
other values were from a spurious incorrect solution. 
OH! It was the first pb that got me. Whoops, thank you so much!