SOLUTION: How would I graph the circle x^2+ y^2-8x+6y+21=0

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Question 951695: How would I graph the circle
x^2+ y^2-8x+6y+21=0
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How would I graph the circle
x^2+ y^2-8x+6y+21=0
Put the equation in center-radius form::
x^2 - 8x + 16 + y^2 + 6y + 9 = -21 + 16 + 9
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(x-4)^2 + (y+3)^2 = 4
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Center:: (4,-3)
Radius:: 2
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Draw a circle with center at (4,-3) and radius = 2
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B+y%5E2-8x%2B6y%2B21=0 to graph it, first write it in standard form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where h and k are x and y coordinates of the center, and r is a radius
%28x%5E2-8x%29%2B+%28y%5E2%2B6y%29%2B21=0....complete squares
%28x%5E2-8x%2B_%29-_%2B+%28y%5E2%2B6y%2B_%29-_%2B21=0
%28x%5E2-8x%2B4%5E2%29-16%2B+%28y%5E2%2B6y%2B3%5E2%29-9%2B21=0
%28x-4%29%5E2%2B+%28y%2B3%29%5E2-25%2B21=0
%28x-4%29%5E2%2B+%28y%2B3%29%5E2-4=0
%28x-4%29%5E2%2B+%28y%2B3%29%5E2=4
so,h=4 and k=-3, the center is at (4,-3)
and r=sqrt%284%29=>r=2
plot a point (4,-3) and draw a circle using r=2