+0  
 
-1
57
1
avatar+216 

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 :)

 Sep 6, 2020
 #1
avatar+27848 
+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

 Sep 7, 2020

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