SOLUTION: I have to write the standard form equation of this circle: Three points on the circle: (-15,-9), (1,-9), and (-15,3).

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Question 945002: I have to write the standard form equation of this circle: Three points on the circle: (-15,-9), (1,-9), and (-15,3).
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
On the coordinate plane, the formula becomes %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
+h and k are the x and y coordinates of the center of the circle , and r is radius
Three points on the circle: (-15,-9), (1,-9), and (-15,3).
use points to find+h, k, and r
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 ......for (-15,-9),
%28-15-h%29%5E2%2B%28-9-k%29%5E2=r%5E2+
h%5E2%2B30h%2Bk%5E2%2B18k%2B306=r%5E2 ............eq.1

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 ......for (1,-9),
h%5E2-2h%2B18k%2Bk%5E2%2B82=r%5E2+ ............eq.2

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 ......for (-15,3),
%28-15-h%29%5E2%2B%283-k%29%5E2=r%5E2+
h%5E2%2B30h%2Bk%5E2-6k%2B234=r%5E2 ............eq.3
system is:
h%5E2%2B30h%2Bk%5E2%2B18k%2B306=r%5E2 ............eq.1
h%5E2-2h%2B18k%2Bk%5E2%2B82=r%5E2+............eq.2
h%5E2%2B30h%2Bk%5E2-6k%2B234=r%5E2 ............eq.3
----------------------------------------------------------
subtract eq.3 from eq.1
h%5E2%2B30h%2Bk%5E2%2B18k%2B306-%28h%5E2%2B30h%2Bk%5E2-6k%2B234%29=r%5E2+-r%5E2
h%5E2%2B30h%2Bk%5E2%2B18k%2B306-h%5E2-30h-k%5E2%2B6k-234=0
18k%2B306%2B6k-234=0
24k%2B72=0
24k=-72
k=-72%2F24
highlight%28k=-3%29
go to
h%5E2-2h%2B18k%2Bk%5E2%2B82=r%5E2+ ............eq.2 substitute k

h%5E2-2h%2B18%28-3%29%2B%28-3%29%5E2%2B82=r%5E2+
h%5E2-2h-54%2B9%2B82=r%5E2+
h%5E2-2h%2B37=r%5E2.............2a
go to
h%5E2%2B30h%2Bk%5E2-6k%2B234=r%5E2 ............eq.3 substitute k and r%5E2
cross%28h%5E2%29%2B30h%2B%28-3%29%5E2-6%28-3%29%2B234=cross%28h%5E2%29-2h%2B37
30h%2B9%2B18%2B234=-2h%2B37
30h%2B261=-2h%2B37
30h%2B2h=-261%2B37
32h=-224
h=-224%2F32
highlight%28h=-7%29
go to
h%5E2-2h%2B18k%2Bk%5E2%2B82=r%5E2+ ............eq.2 substitute k and h and solve for r
%28-7%29%5E2-2%28-7%29%2B18%28-3%29%2B%28-3%29%5E2%2B82=r%5E2+
49%2B14-54%2B9%2B82=r%5E2+
100=r%5E2+
highlight%28r+=10%29

so, your equation is %28x%2B7%29%5E2%2B%28y%2B3%29%5E2=100






Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Her method of substituting the points in the standard form, 
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2,
while correct, is much more difficult than substituting 
in the general form,
x%5E2%2By%5E2%2BDx%2BEy%2BF=0.
I think she used a solver program, so it wasn't hard for her.
But if you have to solve it by hand, like you probably have 
to do, it's much easier to use the general form.

Substituting in the general form,

 

which simplifies to:

system%28-15D-9E%2BF=-306%2CD-9E%2BF=-82%2C+-15D%2B3E%2BF=-234%29  

this is much easier to solve, giving D=14, E=6, and F=-42

and the general equation of the circle is

x%5E2%2By%5E2%2B14x%2B6y-42=0

If you want to get the standard form which she got, you
can complete the squares.  But your teacher will probably
accept the general form.

Edwin