How do you find the radius of a circle if you know three points on the circumference?

Answer 1

First lets look at the equation for a circle... (x - h)2 + (y - k)2 = r2 What do I know? The center must be constant so H and K are two variables I need to solve for... and I know I need to solve for R as well! take your points (i'm making a simple circle with center (0,0) with radius of 1...) 1,0 0,1 -1,0 I have my 3 points.... I will make myself 3 equations: eqn #1a: (1 - h)2 +(0 - k)2 = r2 eqn #2a: (0 - h)2 +(1 - k)2 = r2 eqn #3a: (-1 - h)2 +(0 - k)2 = r2 Ok so now I have some equations to play with... eqn #1b: h2 - 2h +1 + k2 = r2 eqn #2b: h2 + k2-2k +1 = r2 eqn #3b: h2 + 2h +1 + k2= r2 The rest is subsitution... We want r's value so lets see. Lets take eqn #1b and #2b... h2 - 2h +1 + k2 = h2 + k2-2k +1 Balance it out cancle and such try to make it = 0 - 2h = -2k h = k so now I know I can replace h with k or k with h It doesn't matter which one I make to the other if we had 2h = k I would put 2h in its just simpler... keep in mind either way works. eqn #1c: k2 - 2k +1 + k2 = r2 eqn #2c: k2 + k2-2k +1 = r2 eqn #3c: k2 + 2k +1 + k2= r2 collect terms eqn #1c: 2k2 - 2k +1 = r2 eqn 1 is the same as eqn 2! eqn #2c: 2k2 - 2k +1 = r2 eqn #3c: 2k2 + 2k +1= r2 We really have two equations left now it doesn't always work out this way. Set eqn 2 and 3 equal to each other via the r 2k2 - 2k +1 = 2k2 + 2k +1 do some cancling -2k = 2k at this point I know k = 0 its the only viable answer.... take any one of our origonal equations.... h2 - 2h +1 + k2 = r2 all the parts with k or h = 0; remember h = k = 0 1 = r2 1 = r If I had time I would do a more complex equation for yea... but this kinda gets the point across...