+218 I know this was already asked here: https://web2.0calc.com/questions/a-circle-passes-through-the-points-2-0-2-0-and-3 but I could not figure out the rest of the problem. This is what I had done:
(h-(-2))^2 + (k-0)^2 = r^2
(h-2)^2 + (k-0)^2 = r^2
(h-3)^2 + (k-2)^2 = r^2
(h-2)^2 + (h+2)^2 + (k^2) - (k^2) = 0
(h-3)^2 + (k-2)^2 - (k^2) = 0
h^2 + 4h + 4 - h^2 + 4h + 4 + k^2 - k^2 = 0
h^2 - 6h + 9 - h^2 + 4h + 4 + k^2 - 4k - 4 - k^2 = 0
h= -1
k=11/4
Thanks!
Noori :)
+37159 I think you have the h,k backwards in your parentheses....
first equation (-2-h)^2 + (0-k)^2 = r2
second equatio (2-h)^2 + (0- k)^2 = r^2
third equation (3-h)^2 + (2-k)^2 = r^2
Solve for h k r
