Question

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?

Answer 1

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

(y-110) = (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

(y-20) = -5(x-100)
y = -5x +520

The two pb's intersect at (50,270)

so, it looks like the circle is

(x-30)^2 + (y+30)^2 = 75400

sqrt(75400) = 274.59 or about 275

diameter would thus be 550.

Answer 2

Oops. The equation is
(x-50)^2 + (y+-270)^2 = 75400

other values were from a spurious incorrect solution.

Answer 3

OH! It was the first pb that got me. Whoops, thank you so much!