SOLUTION: i have to find the center of a circle with an equation (x-2)squared+(y+3)squared=16

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Question 97513: i have to find the center of a circle with an equation (x-2)squared+(y+3)squared=16
Found 2 solutions by mathslover, edjones:
Answer by mathslover(132) About Me  (Show Source):
You can put this solution on YOUR website!

the general equation of a circle is
%28x-x1%29%5E2+%2B+%28y-y1%29%5E2+=+r%5E2+ where (x1,y1) is the center of the circle and
r its radius
given
equation of the circle
+%28x-2%29%5E2+%2B+%28y+%2B+3%29%5E2+=+16+
rewriting this equation in the standard form
+%28x-2%29%5E2+%2B+%28y+-+%28-3%29%29%5E2+=+4%5E2+
clearly the centre is (2,-3) and the radius 4

Answer by edjones(7311) About Me  (Show Source):
You can put this solution on YOUR website!
(x-h)^2+(y-k)^2=r^2
(x-2)^2+(y+3)^2=16
h=2 k=-3 (2,-3)
Ed
drawing%28500%2C500%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2Ccircle%282%2C-3%2C4%29%29