SOLUTION: Write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0). Show all work to receive credit

Algebra ->  Points-lines-and-rays -> SOLUTION: Write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0). Show all work to receive credit      Log On


   

Question 316948: Write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0). Show all work to receive credit
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The general equation for a circle centered at (h,k) with a radius R is:
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2
or more usefully,
%28h-x%29%5E2%2B%28y-k%29%5E2=R%5E2
J:%28h%2B6%29%5E2%2B%28k-0%29%5E2=R%5E2
K:%28h%2B3%29%5E2%2B%28k-3%29%5E2=R%5E2
L:%28h-0%29%5E2%2B%28k-0%29%5E2=R%5E2
From the L equation,
h%5E2%2Bk%5E2=R%5E2
From the J equation,
%28h%2B6%29%5E2%2Bk%5E2=R%5E2
Equate those two,
h%5E2%2Bk%5E2=%28h%2B6%29%5E2%2Bk%5E2
h%5E2=%28h%2B6%29%5E2
h%5E2=h%5E2%2B12h%2B36
12h%2B36=0
highlight%28h=-3%29
From the K equation,
%28k-3%29%5E2=R%5E2
From the L equation,
9%2Bk%5E2=R%5E2
Equate those two,
9%2Bk%5E2=k%5E2-6k%2B9
-6k=0
highlight%28k=0%29
Finally using L,
9%2B0=R%5E2
highlight%28R%5E2=9%29
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highlight_green%28%28x%2B3%29%5E2%2By%5E2=9%29