SOLUTION: Find an equation of the circle whose center is at the point (-3 , 6) and is tangent to the y axis. I figure the answer would be (x+3)^2+(y-6)^2=9 but however at this moment the

Algebra ->  Coordinate-system -> SOLUTION: Find an equation of the circle whose center is at the point (-3 , 6) and is tangent to the y axis. I figure the answer would be (x+3)^2+(y-6)^2=9 but however at this moment the      Log On


   

Question 1088136: Find an equation of the circle whose center is at the point (-3 , 6) and is tangent to the y axis.
I figure the answer would be (x+3)^2+(y-6)^2=9 but however at this moment there isn't anyway to prove the answer to whether to be correct or wrong. Thank you very much!
Found 2 solutions by ikleyn, Fombitz:
Answer by ikleyn(52943) About Me  (Show Source):
You can put this solution on YOUR website!
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%28x%2B3%29%5E2%2B%28y-6%29%5E2 = 3%5E2,

or

%28x%2B3%29%5E2%2B%28y-6%29%5E2 = 9.



Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use the general equation of a circle of radius R centered at (h,k),
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2
%28x%2B3%29%5E2%2B%28y-6%29%5E2=R%5E2
Since it's tangent to the y-axis, (0,6) is also on the circle,
%280%2B3%29%5E2%2B%286-6%29%5E2=R%5E2
R%5E2=9
So,
%28x%2B3%29%5E2%2B%28y-6%29%5E2=9
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