Question

i am trying to find the center of a circle using these points (6,13)(5,12)(14,9)

Answer 1

define a general point (a,b)

and work out the d^2 = (x-a)^2 + (y-b)^2 for each given point

and then equate them all to determine a and b as the center of the circle

Answer 2

I am not sure how to get the r2

Answer 3

r is just the distance from a,b to any of the points given

Answer 4

I guess I am just lost can you show me

Answer 5

d^2 = (x-a)^2 + (y-b)^2

d^2 = x^2 -2ax +a^2 + y^2 -2by + b^2

using the points (6,13) (5,12) (14,9)

d^2 = 6^2 -2a6 +a^2 + 13^2 -2b13 + b^2
d^2 = 5^2 -2a5 +a^2 + 12^2 -2b12 + b^2
d^2 = 14^2 -2a14 +a^2 + 9^2 -2b9 + b^2

equating these should help solve for a,b .... and once a,b is known the r^2 is known

Answer 6

What about bisecting each chord, and finding where the bisectors intersect?

Answer 7

ow my head!! yeah thatll work too if you wanna take that route :)

Answer 8

http://mathforum.org/library/drmath/view/55239.html

my idea seems to be matrix applicable :)

Answer 9

not too bad.
slope between (5,12) to (6,13) is 1, and perpendicular is -1 and goes through (5.5,12.5)
y-12.5 = -(x-5.5) is the first bisector

slope between (6,13) to (14,9) is (13-9)/(6-14)= 4/-8= -1/2
slope of perpendicular is +2 through pt (10,11)
y-11= 2(x-10) is the 2nd bisector

not too hard to find the intersection.

Answer 10

@phi, I like that.

Answer 11

pfft ... yeah, if you like simple ;)