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 :)

Noori Sep 6, 2020

#1**+1 **

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

ElectricPavlov Sep 7, 2020